maxima and minima of f(x,y,z)

The question: find the point on the plane $\displaystyle x-y+z=4$ that is closest to the point $\displaystyle (1,2,3)$.

So I guess in math terms the problem is to find $\displaystyle min(d)$ where

$\displaystyle d=\sqrt{(x-1)^{2}+(y-2)^{2}+(z-3)^{2}}$

It's clear that

$\displaystyle d_{x}=\frac{x-1}{\sqrt{(x-1)^{2}+(y-2)^{2}+(z-3)^{2}}}$

$\displaystyle d_{y}=\frac{y-2}{\sqrt{(x-1)^{2}+(y-2)^{2}+(z-3)^{2}}}$

$\displaystyle d_{z}=\frac{z-3}{\sqrt{(x-1)^{2}+(y-2)^{2}+(z-3)^{2}}}$

And of course, what you're supposed to do is set them equal to zero and solve. Only problem is, I have no clue how to do that. Any help appreciated. Thanks in advance.

Edit: Never mind, is it as stupid as I think it is?

$\displaystyle d_{x}=0\ \Rightarrow\ x-1=0\ \Rightarrow\ x=1$

Is that the solution? If so, I think I should bash my head against a wall.