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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

beginalign*
d &= sqrt(x-8)^2 + (y-0)^2 + (-x-y+12)^2 \
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
endalign*

Find partial derivative $f_x$ and $f_y$ and critical points

beginalign*
f_x &= 2(x-8) + 2(-x-y+12) \
&= 24-2y quad (textset = 0) \
&= textcritical point y = 4 \
f_y &= 2y + 2(-x-y+12) \
&= 24 - 2x quad (textset = 0) \
&= textcritical point x = 12 \
endalign*

Plug in $x = 12$ and $y = 4$ to original equation

$$d = sqrt(x - 8)^2 + (4)^2 + (-12-4+12)^2 = sqrt48$$

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 $fracpartialpartial x left[ (x-8)^2+y^2+(-x-y+12)^2 right]$ 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.

The distance formula is $$D=fracax_0+by_0 +cz_0-dsqrta^2+b^2+c^2=frac4sqrt3.$$

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 $frac43$ to each component so that $x+y+z$ now equals $6$. Therefore, the distance is $|(frac43, frac43, frac43)|=sqrt3frac169=sqrtfrac163=frac4sqrt3$

It's $$fracsqrt1^2+1^2+1^2=frac4sqrt3.$$
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
$$fracsqrta^2+b^2+c^2.$$

Also, we can end your way.

Indeed, by C-S
$$sqrt(x-8)^2+y^2+(z+6)^2=frac1sqrt3sqrt(1^2+1^2+1^2)((x-8)^2+y^2+(z+6)^2)geq$$
$$geqfrac1sqrt3sqrt(x-8+y+z+6)^2=frac4sqrt3$$ and we got the distance again.

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?).

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.