<s>[INST] $ax+by+cz=d$ is the equation of a plane in space. Show that $d'$ is the distance from the plane to the origin. This is a 3 part practice question I would like to get some feedback on. I think I have solved the 1st two parts, but I need a little direction for part (c) (the title is Part (a) ) which is repeated here with more detail,
a) $ax+by+cz=d$ is the equation of a plane in space.  Divide this equation by a real number to get $a'x+b'y+c'z=d'$. 
Show that $d'$ is the distance from the plane to the origin.
It is required that $ a'^2 + b'^2 + c'^2 = 1 $ and $d' >= 0$
Here is my solution to part (a)
Let $p$ be a real number $ (p \neq 0)$, such that
$\frac{1}{p}(ax + by + cz) = \frac{d}{p} = a'x + b'y + c'z = d'$
It is required that 
$ a'^2 + b'^2 + c'^2 = 1 $  so,
$ (\frac{a}{p})^2 + (\frac{b}{p})^2 + (\frac{c}{p})^2 = 1 $, this implies
$ a^2 + b^2 + c^2 = p^2 $
$ \pm\sqrt{a^2 + b^2 + c^2} =  p $
Since $d = ax_0 + by_0 + cz_0 $ (where $ Q(x_0,y_0,z_0) $ is a point on the plane),
$d' = \frac{1}{p}(ax_0 + by_0 + cz_0)$
Substituting $ p $ into $d'$, we get
$d' = \frac{1}{\sqrt{a^2 + b^2 + c^2}}(ax_0 + by_0 + cz_0) = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}}$ 
(taking the positive since it is required that $d' >= o$)
Since 
$ ai + bj + ck $ is a normal vector to the plane and,
$ x_0i + y_0j + z_0k $ is the position vector of point $Q$
$d' = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}} = \frac{|n \cdot \vec{\mathbf{OQ}|} }{|n|}$ 
This is the formula from a point to a plane (in this case the origin).
b) Find the position vector of the point $S$ that is closest to the origin on the plane, with equation in the primed form of part (a).
My solution to part (b):
From part (a) it was determined that  
$d' = \frac{ax_0 + by_0 + cz_0}{\sqrt{a^2 + b^2 + c^2}} = \frac{|n \cdot \vec{\mathbf{OQ}|} }{|n|}$ , is the distance from the plane to the origin. 
And, in the primed form
$ \vec{\mathbf{n}} = a'i + b'j + c'k $ is a normal vector to the plane
Since $ \vec{\mathbf{n}} \cdot \vec{\mathbf{n}} = |\vec{\mathbf{n}}|^2  =  a'^2 + b'^2 + c'^2 = 1 $ so $ |\vec{\mathbf{n}}| = 1$
$ \vec{\mathbf{n}} $ is the unit normal vector to  $ a'x + b'y + c'z = d'$
The position vector $ \vec{\mathbf{S}} $ is then:
$ d'\vec{\mathbf{n}} = d' (a'i + b'j + c'k) $
Is this a sufficient answer? Or should I expand d'?
c) Consider the plane given by the equation in the primed form of part (b). Why is each plane containing the origin described by two distinct equations of this
form? Why does each plane that does not contain the origin have a unique equation
of this form?
This question, I need a little direction.
If the equation of a plane in form $ a'x + b'y + c'z = d'$  contains the origin, then
$d' = 0$, so 
$ a'x + b'y + c'z =  \frac{a}{p}x + \frac{b}{p}x + \frac{c}{p}x = 0$
What would be a 2nd distinct equation of this form? Could someone point me in the right direction?
 [/INST] Your answer looks good. Here is an alternate method to find the minimum point.
To minimize
$$
x^2+y^2+z^2\tag{1}
$$
while maintining
$$
ax+by+cz=d\tag{2}
$$
we want to have
$$
2x\,\delta x+2y\,\delta y+2z\,\delta z=0\tag{3}
$$
for all $(\delta x,\delta y,\delta z)$ so that
$$
a\,\delta x+b\,\delta y+c\,\delta z=0\tag{4}
$$
By orthogonality, this means we need $(a,b,c)\|(x,y,z)$; that is,
$$
(x,y,z)=k(a,b,c)\tag{5}
$$
Substituting $(5)$ into $(2)$ yields
$$
k(a^2+b^2+c^2)=d\tag{6}
$$
Plugging $(5)$ and $(6)$ into $(1)$, we get the square of the minimum distance is
$$
\left(\frac{d}{a^2+b^2+c^2}\right)^2(a^2+b^2+c^2)\tag{7}
$$
Thus, the minimum distance is
$$
\frac{|d|}{\sqrt{a^2+b^2+c^2}}\tag{8}
$$
</s>