<s>[INST] How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate. 
The following is my work: 
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$$
since $x+y+z = 6$, $z = 6-x-y$, so 
\begin{align*}
d &= \sqrt{(x-8)^2 + (y-0)^2 + (-x-y+12)^2} \\
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
\end{align*}
Find partial derivative $f_x$ and $f_y$ and critical points
\begin{align*}
f_x &= 2(x-8) + 2(-x-y+12) \\
&= 24-2y \quad (\text{set }= 0) \\
&= \text{critical point }y = 4 \\
f_y &= 2y + 2(-x-y+12) \\
&= 24 - 2x \quad (\text{set }= 0) \\
&= \text{critical point }x = 12 \\
\end{align*}
Plug in $x = 12$ and $y = 4$ to original equation 
$$d = \sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = \sqrt{48}$$
 [/INST] Your approach is a good one, and in fact can lead to the general formula mentioned in the other answers.
You just made a few mistakes in your calculation.
For example, you say that $\frac{\partial}{\partial x} \left[ (x-8)^2+y^2+(-x-y+12)^2 \right]$ is $2(x-8)+2(-x-y+12)$, when really it is $2(x-8)-2(-x-y-12)$.  You forgot the chain rule.
You make a similar chain rule mistake when calculating the partial derivative with respect to $y$.
I think if you fix these, you should be able to finish your calculation the way you started it.
Note to other posters:  I think it is generally of greater pedagogical value to help a student realize how to fix their own reasoning, rather than supplying them with a totally different method.  Our primary aim should be to empower students to realize that they have the power to think through these things logically and creatively.
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