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As you know, the electrostatic potential for a point charge in free space (at the position $\vec {r_0}=x_0\hat {x}+y_0\hat{y}+z_0\hat{z}$) is given by the relation:

$$V(\vec r)=\frac{1}{4\pi\epsilon_0}\frac{q}{\left|\vec r-\vec {r_0}\right|}$$

(I have assumed to assign zero volt to infinite distance from the source).

I want try to derive it by using this expression of volume charge density:

$$\rho(\vec r)=q \cdot \delta (x-x_0)\delta (y-y_0)\delta (z-z_0).$$

The solution of Poisson equation is given by:

$$V(\vec {r}_0)=\int _{Volume} \frac{1}{4\pi\epsilon_0 |{\vec r-\vec {r}_0}|}\rho(r)dV$$

Therefore:

$$V=\int _{Volume} \frac{1}{4\pi\epsilon_0 |(x-x_0)\hat {x}+(y-y_0)\hat{y}+(z-z_0)\hat{z}|}q \cdot \delta (x-x_0)\delta (y-y_0)\delta (z-z_0)dV$$

If we apply the property of dirac delta function, we will get infinite. How do we solve this problem?

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You need to make a distinction between the integrating variables and the input to the potential function. If you want to find the potential at the point $(x,y,z)$, then you need to integrate over the charge distribution at all points $(x',y',z')$. So your expression should actually look like:

$$V(x,y,z)=\int \frac{1}{4\pi\epsilon_0|(x-x')\hat{x}+(y-y')\hat{y}+(z-z')\hat{z}|}q\cdot\delta(x'-x_0)\delta(y'-y_0)\delta(z'-z_0)dx'dy'dz'$$

When you integrate, the Dirac delta functions substitute $x_0$ for $x'$ in the expression, so you get

$$V(x,y,z)=\frac{q}{4\pi\epsilon_0|(x-x_0)\hat{x}+(y-y_0)\hat{y}+(z-z_0)\hat{z}|}$$

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You seem to confuse the position of the point charge $r_0$ with the argument of the potential in Poissons equation. In fact,

$$V(\vec{r}')=\int _{Volume} \frac{1}{4\pi\epsilon_0 |{\vec r-\vec{r}'}|}\rho(\vec{r})dV$$

Plugging in $\rho(r) = q\delta(\vec{r}-\vec{r_0}) $ immediately yields

$$V(\vec{r}')= \frac{q}{4\pi\epsilon_0 |{\vec{r_0}-\vec r'}|}$$

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