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 
$$d = \sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = \sqrt{48}$$
 A: 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.
A: The distance formula is $$D=\frac{|ax_0+by_0 +cz_0-d|}{\sqrt{a^2+b^2+c^2}}=\frac{4}{\sqrt3}.$$
A: The closest distance will be along a line perpendicular to the plane or parallel to $(1,1,1)$. Since $8+0+-6=2$ we must add $\frac{4}{3}$ to each component so that $x+y+z$ now equals $6$. Therefore, the distance is $|(\frac{4}{3}, \frac{4}{3}, \frac{4}{3})|=\sqrt{3\frac{16}{9}}=\sqrt{\frac{16}{3}}=\frac{4}{\sqrt{3}}$
A: It's $$\frac{|8+0-6-6|}{\sqrt{1^2+1^2+1^2}}=\frac{4}{\sqrt3}.$$
I used the following formula.
The distance from the point $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ it's
$$\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.$$
Also, we can end your way.
Indeed, by C-S
$$\sqrt{(x-8)^2+y^2+(z+6)^2}=\frac{1}{\sqrt3}\sqrt{(1^2+1^2+1^2)((x-8)^2+y^2+(z+6)^2)}\geq$$
$$\geq\frac{1}{\sqrt3}\sqrt{(x-8+y+z+6)^2}=\frac{4}{\sqrt3}$$ and we got the distance again.
A: Option:
Normal of the plane: $(1,1,1)$, 
Normalized(i.e of unit length): $(1/√3)(1,1,1)$
Line through $(8,0,-6)$:
$\vec r= (8,0,-6)+t (1/√3)(1,1,1)$.
Determine $t$ when line intersects plane:
$(8+t/√3)+t/√3+(-6+t/√3)=6;$
$3(t/√3)=4$;
$t= 4/√3.$
Distance = $4/√3$ (Why?).
A: You can draw a line from your point to the plane and then find the length of its projection onto the unit normal vector to the plane.
