From the point (2, -1, 3) to the plane z - 2x = 3
We have the distance from $\displaystyle (x,y,z) $ to $\displaystyle (2,-1,3) $ is given by $\displaystyle d=\sqrt{(x-2)^2+(y+1)^2+(z-3)^2} $
We know though that $\displaystyle z=2x+3 $ and also the critical points of $\displaystyle d $ will be the same as the critical points of $\displaystyle d^2 $. So let's minimize $\displaystyle d^2 $ instead.
$\displaystyle d^2 = f(x,y) = (x-2)^2+(y+1)^2+4x^2 $
Now we solve $\displaystyle \frac{\partial f}{\partial x} = 2(x-2)+8x=0 $ and $\displaystyle \frac{\partial f}{\partial x} = 2(y+1)=0 $.
Thus we get one critical point, namely $\displaystyle (x,y)=\left(\frac25,-1\right) $.
I'll leave it to you to verify it minimizes the distance. After that, plug $\displaystyle (x,y) $ into $\displaystyle d $ and you'll have your answer.
$\displaystyle D = \hat{n} \cdot (\vec{p} - \vec{q})$ where $\displaystyle \hat{n}$ is a unit normal vector to the plane, $\displaystyle \vec{p}$ is the position vector of a known point in the plane and $\displaystyle \vec{q}$ is the position vector of the point from which you're calculating the distance from.