From the point (2, -1, 3) to the plane z - 2x = 3
We have the distance from to is given by
We know though that and also the critical points of will be the same as the critical points of . So let's minimize instead.
Now we solve and .
Thus we get one critical point, namely .
I'll leave it to you to verify it minimizes the distance. After that, plug into and you'll have your answer.