As usual let $|l,m\rangle$ denote an eigenstate of $\vec{L}^2$ and $L_z$. I know that

\begin{align} \vec{L}^2 |l,m\rangle &= \hbar^2 l(l+1) |l,m\rangle, \\ L_z |l,m\rangle &= \hbar m |l,m\rangle. \end{align}

Consider $|1,1\rangle$. Then, I can write $|1,1\rangle$ in terms of $|1,m_x\rangle$ with $m_x = \pm 1,0$, where $|1,m_x\rangle$ denotes an eigenbasis of $L_x$, as

\begin{align} |1,1\rangle_z = c_1\cdot |1,-1\rangle_x + c_2 \cdot |1,0\rangle_x + c_3 \cdot |1,1\rangle_x. \end{align}

I want to compute the coefficients $c_1, c_2, c_3$ working with

\begin{align} \Delta^2L_x := \langle (L_x - \langle L_x \rangle)^2\rangle = \frac{1}{2} \cdot \left( \langle L^2\rangle - \langle L_z^2\rangle\right) = \frac{\hbar^2}{2}\cdot\left(l(l+1) - m^2\right) \end{align}

as in this case (for $|1,1\rangle$) we have $\Delta^2L_x = \frac{\hbar^2}{2}$. However, I don't understand how this can be done. Can someone please explain it?