Your Conclusion:
$q_{1}, q_{2}$ must have same polarity and $q_{3}$ must have different polarity.
is wrong by itself : If the triad of charges $\,\left(q_{1}, q_{2},q_{3}\right)\,$ is in equilibrium if and only if $\,\lbrace q_{1}\cdot q_{2}>0 \,\land\, q_{1}\cdot q_{3}<0\rbrace\,$ then the triad of charges which comes out from
their permutation $\,\left(q'_{1}\!=\!q_{1}, q'_{2}\!=\!q_{3},q'_{3}\!=\!q_{2}\right)\,$ is also in equilibrium but violates this rule since $\,\lbrace q'_{1}\cdot q'_{2}\!=\!q_{1}\cdot q_{3}<0 \,\land \, q'_{1}\cdot q'_{3}\!=\!q_{1}\cdot q_{2}>0\rbrace$.
Hint : Without loss of generality fix the charges $\,q_{1}\,$ and $\,q_{2}\,$ on the $\,x-$axis at a given distance $\,d\,$ apart and place the charge $\,q_{3}\,$
on the same axis at coordinate $\,x\,$ to be determined for the system to be in equilibrium.
Try to prove that the coordinate $\,x\,$ must satisfy the equation
\begin{equation}
\dfrac{\dfrac{x\!-\!d\hphantom{^{3}}}{\vert x\!-\!d\vert^{3}}}{\dfrac{x\hphantom{^{3}}}{\vert x\vert^{3}}}=-\dfrac{q_{1}}{q_{2}}
\tag{Hint-A}
\end{equation}
or
\begin{equation}
\dfrac{\xi\hphantom{^{3}}}{\vert \xi\vert^{3}}=-\dfrac{q_{1}}{q_{2}} \, \quad \text{where} \:\:\xi \equiv \dfrac{x\!-\!d}{x}
\tag{Hint-A$^{\prime}$}
\end{equation}
and the charge $\,q_{3}\,$ the equation
\begin{equation}
q_{3}=-\dfrac{\vert x\vert^{3}}{x\hphantom{^{3}}}\dfrac{q_{2}}{d^{2}}=\dfrac{\vert x\!-\!d\vert^{3}}{x\!-\!d\hphantom{^{3}}}\dfrac{q_{1}}{d^{2}}
\tag{Hint-B}
\end{equation}
where in (Hint-B) $\,x\,$ is a solution of (Hint-A).
EDIT
We'll use the following symbols :
\begin{align}
\mathbf{f}_{ji} & =\text{force on charge $\,q_{i}\,$ from charge $\,q_{j}\,$}\, , \quad j\ne i
\tag{symbols-01}\\
\mathbf{F}_{i} & =\sum\limits_{j\ne i} \mathbf{f}_{ji}=\text{resultant force on charge $\,q_{i}$}
\tag{symbols-02}
\end{align}
Obviously the problem is 1-dimensional, for if the three charges don't lie on a line then for the force on charge $\,q_{i}\,$ we would have $\,\mathbf{F}_{i}\ne \boldsymbol{0}\,$ as the resultant of the two non-collinear forces $\,\mathbf{f}_{ji}\, ,j\ne i\,,$ from the other two charges.
So let the three charges $\,q_{1},q_{2},q_{3}\,$ on the $\,x\!-\!$axis as in Figure 01. For the forces on the three charges we have (note that $\,\mathbf{f}_{ji}=-\mathbf{f}_{ij}$ and $\,x \ne0,d\,$)
\begin{align}
q_{1} : \mathbf{f}_{21} & =-k\dfrac{q_{1}q_{2}}{d^{2}}\, , \quad \mathbf{f}_{31} =-k\dfrac{q_{1}q_{3}}{\vert x \vert ^{3}}x
\tag{01a}\\
\mathbf{F}_{1} & =\mathbf{f}_{21}+\mathbf{f}_{31} =-k\dfrac{q_{1}q_{2}}{d^{2}}-k\dfrac{q_{1}q_{3}}{\vert x \vert ^{3}}x = -kq_{1}\left[\dfrac{q_{2}}{d^{2}}+\dfrac{q_{3}}{\vert x \vert ^{3}}x \right]
\tag{01b}\\
q_{2} : \mathbf{f}_{12} & =+k\dfrac{q_{2}q_{1}}{d^{2}}\, , \quad \mathbf{f}_{32} =-k\dfrac{q_{2}q_{3}}{\vert x\!-\!d \vert ^{3}}\left(x\!-\!d\right)
\tag{02a}\\
\mathbf{F}_{2} & =\mathbf{f}_{12}+\mathbf{f}_{32} =+k\dfrac{q_{2}q_{1}}{d^{2}}-k\dfrac{q_{2}q_{3}}{\vert x\!-\!d \vert ^{3}}\left(x\!-\!d\right)=+k q_{2}\left[\dfrac{q_{1}}{d^{2}}-\dfrac{q_{3}}{\vert x\!-\!d \vert ^{3}}\left(x\!-\!d\right)\right]
\tag{02b}\\
q_{3} :\mathbf{f}_{13} & =+k\dfrac{q_{3}q_{1}}{\vert x \vert ^{3}}x \, , \quad \mathbf{f}_{23} =+k\dfrac{q_{3}q_{2}}{\vert x\!-\!d \vert ^{3}}\left(x\!-\!d\right)
\tag{03a}\\
\mathbf{F}_{3} & =\mathbf{f}_{13}+\mathbf{f}_{23} =+k\dfrac{q_{3}q_{1}}{\vert x \vert ^{3}}x+k\dfrac{q_{3}q_{2}}{\vert x\!-\!d \vert ^{3}}\left(x\!-\!d\right)
=+kq_{3}\left[\dfrac{q_{1}}{\vert x \vert ^{3}}x+\dfrac{q_{2}}{\vert x\!-\!d \vert ^{3}}\left(x\!-\!d\right)\right]
\tag{03b}
\end{align}
Note that as expected (no external forces, zero sum of internal forces)
\begin{equation}
\mathbf{F}_{1}\!+\!\mathbf{F}_{2}\!+\!\mathbf{F}_{3}=\boldsymbol{0}
\tag{04}
\end{equation}
so if we set
\begin{equation}
\mathbf{F}_{1}=\boldsymbol{0}\, , \quad \mathbf{F}_{2}=\boldsymbol{0}
\tag{05}
\end{equation}
then automatically
\begin{equation}
\mathbf{F}_{3}=\boldsymbol{0}
\tag{06}
\end{equation}
and the system is in equilibrium.
Now, for (05) to be valid we must have from (02b) and (01b) respectively
\begin{align}
\dfrac{x\!-\!d\hphantom{^{3}}}{\vert x\!-\!d\vert^{3}} & =\hphantom{-}\dfrac{1}{d^{2}}\dfrac{q_{1}}{q_{3}}
\tag{07a}\\
\dfrac{x\hphantom{^{3}}}{\vert x\vert^{3}} \hphantom{\:\:\;} & =-\dfrac{1}{d^{2}}\dfrac{q_{2}}{q_{3}}
\tag{07b}
\end{align}
Dividing equations (07) yields
\begin{equation}
\dfrac{\dfrac{x\!-\!d\hphantom{^{3}}}{\vert x\!-\!d\vert^{3}}}{\dfrac{x\hphantom{^{3}}}{\vert x\vert^{3}}}=-\dfrac{q_{1}}{q_{2}}
\tag{08}
\end{equation}
that is the hint equation (Hint-A), or
\begin{equation}
\dfrac{\xi\hphantom{^{3}}}{\vert \xi\vert^{3}}=-\dfrac{q_{1}}{q_{2}} \, \quad \text{where} \:\:\xi \equiv \dfrac{x\!-\!d}{x}
\tag{09}
\end{equation}
that is the hint equation (Hint-A$^{\prime}$).
We note that equation (08) is independent of the value of the charge $\,q_{3}$. This might be expected, since its solution with respect to $\,x\,$ give us the point where the resultant electrostatic field of charges $\,q_{1},q_{2}\,$ is zero. In other words, on this point $\,\mathbf{F}_{3}= \boldsymbol{0}\,$, see equation (03b).
To the contrary, for the forces $\,\mathbf{F}_{1},\mathbf{F}_{2}\,$ to be both zero the value of $\,q_{3\,}$ must satisfy the equation
\begin{equation}
\bbox[#FFFF88,5px,border:1px solid black]{
-\dfrac{\vert x\vert^{3}}{x\hphantom{^{3}}}\dfrac{q_{2}}{d^{2}}=q_{3}=\dfrac{\vert x\!-\!d\vert^{3}}{x\!-\!d\hphantom{^{3}}}\dfrac{q_{1}}{d^{2}}}
\tag{10}
\end{equation}
that is the hint equation (Hint-B).
Note that equation (10), since it includes equation (08), is sufficient to give us the values of $\,x,q_{3}\,$ in order to have system equilibrium.
From (09) we have
\begin{equation}
\vert \xi\vert^{2}=\dfrac{\vert q_{2}\vert}{\vert q_{1}\vert} \hphantom{\:\:} \Longrightarrow \hphantom{\:\:} \xi=\pm \sqrt{\dfrac{\vert q_{2}\vert}{\vert q_{1}\vert}}\hphantom{\:\:} \Longrightarrow \hphantom{\:\:} \dfrac{x\!-\!d}{x}=\text{sign}\left(\xi\right)\sqrt{\dfrac{\vert q_{2} \vert}{\vert q_{1} \vert}}
\tag{11}
\end{equation}
where
\begin{equation}
\text{sign}\left[\xi\right]\equiv \text{sign of $\,\xi\,$}\stackrel{\text{(09)}}{=\!\!=\!\!=}\text{sign}\left[\!-\dfrac{q_{1}}{q_{2}}\right]=-\text{sign}\left[\dfrac{q_{1}}{q_{2}}\right]
\tag{12}
\end{equation}
and so (11) yields
\begin{equation}
\left(\!1\!+\!\text{sign}\left[\dfrac{q_{1}}{q_{2}}\right]\sqrt{\dfrac{\vert q_{2}\vert}{\vert q_{1}\vert}}\:\right)x = d
\tag{13}
\end{equation}
Equation (13) has the solution
\begin{equation}
x=\dfrac{d}{\left(\!1\!+\!\text{sign}\left[\dfrac{q_{1}}{q_{2}}\right]\sqrt{\dfrac{\vert q_{2}\vert}{\vert q_{1}\vert}}\:\right)} \qquad \textbf{if} \qquad \left(\!1\!+\!\text{sign}\left[\dfrac{q_{1}}{q_{2}}\right]\sqrt{\dfrac{\vert q_{2}\vert}{\vert q_{1}\vert}}\:\right)\ne 0
\tag{14}
\end{equation}
while the problem is impossible if
\begin{equation}
1\!+\!\text{sign}\left[\dfrac{q_{1}}{q_{2}}\right]\sqrt{\dfrac{\vert q_{2}\vert}{\vert q_{1}\vert}}=0
\tag{15}
\end{equation}
Above equation is valid if and only if $\,q_{2}=-q_{1}\,$ so we have as a first case :
\begin{equation}
\bbox[#FFFFF0,5px,border:1px solid black]{
\textbf{CASE A :} \qquad q_{2}=-q_{1} \:\:\:\Longrightarrow \:\:\:\text{problem impossible}}
\tag{16}
\end{equation}
Now let proceed to other CASES. Suppose that the two charges are both positive or both negative, that is $\,q_{1}q_{2}>0\,$. Then from (14)
\begin{equation}
x=\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!+\!\sqrt{\vert q_{2} \vert}}\,d
\tag{17}
\end{equation}
and from (10)
\begin{equation}
-\left(\!\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!+\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}q_{2}=q_{3}=-\left(\!\dfrac{\sqrt{\vert q_{2} \vert}}{\sqrt{\vert q_{1} \vert}\!+\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}q_{1}
\tag{18}
\end{equation}
Note that the charge $\,q_{3}\,$ must be placed between the two charges $\,q_{1},q_{2}\,$ having charge opposite to the common charge of them.
So
\begin{equation}
\bbox[#FFFFF0,5px,border:1px solid black]{
\textbf{CASE B :} \quad q_{1}q_{2}>0 \Longrightarrow
\begin{cases}
x\hphantom{_{3}}=\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!+\!\sqrt{\vert q_{2} \vert}}\,d \\
\\
q_{3}=-\left(\!\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!+\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}q_{2}=-\left(\!\dfrac{\sqrt{\vert q_{2} \vert}}{\sqrt{\vert q_{1} \vert}\!+\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}q_{1}
\end{cases}}
\tag{19}
\end{equation}
see also Figure 02 at the end(1).
For the next CASE C suppose that $\,q_{1}q_{2}<0\,$, that is one is positive and the other negative. To exclude the CASE A, see equation (16), we further assume $\,q_{2}\ne -q_{1}$. Then from (14)
\begin{equation}
x=\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\,d
\tag{20}
\end{equation}
Now
\begin{align}
\text{If} \:\vert q_{1} \vert > \vert q_{2} \vert & \Longrightarrow x>\hphantom{\!\!-}d>0 \stackrel{\text{(10)}}{ \Longrightarrow } q_{3}=-\left(\!\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}\!q_{2}=+\left(\!\dfrac{\sqrt{\vert q_{2} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}\!q_{1}
\tag{21a}\\
\text{If} \:\vert q_{1} \vert < \vert q_{2} \vert & \Longrightarrow x<\!\!-d<0 \stackrel{\text{(10)}}{ \Longrightarrow } q_{3}=+\left(\!\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}\!q_{2}=-\left(\!\dfrac{\sqrt{\vert q_{2} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}\!q_{1}
\tag{21b}
\end{align}
So CASE C could be summarized as follows
\begin{equation}
\bbox[#FFFFF0,5px,border:1px solid black]{
\textbf{CASE C :} \quad
\begin{vmatrix}
q_{1}q_{2}<0 \\
q_{2}\ne -q_{1}
\end{vmatrix}
\Longrightarrow
\begin{cases}
x\hphantom{_{3}}=\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\,d \\
\\
q_{3}=
\begin{cases}
-\left(\!\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}\!q_{2}\quad \text{if} \quad \vert q_{1} \vert > \vert q_{2} \vert \\
+\left(\!\dfrac{\sqrt{\vert q_{1} \vert}}{\sqrt{\vert q_{1} \vert}\!-\!\sqrt{\vert q_{2} \vert}}\right)^{\!\!\!2}\!q_{2}\quad \text{if} \quad \vert q_{1} \vert < \vert q_{2} \vert
\end{cases}
\end{cases}}
\tag{22}
\end{equation}
(1)
In the Figures below the charges are shown as vertical vectors parallel to the $y-$axis. The positives towards the $+y$ in red color, the negatives towards the $-y$ in blue color.
