<s>[INST] Find the shortest distance between the point $(8,3,2)$ and the line through the points $(1,2,1)$ and $(0,4,0)$ "Find the shortest distance between the point $(8,3,2)$ and the line through the points $(1,2,1)$ and $(0,4,0)$"
$$P = (1,2,1), Q = (0,4,0), A = (8,3,2)$$
$OP$ = vector to $P$
$$PQ_ = (0,4,0) - (1,2,1)$$
I found that the equation of the line $L$ that passes through $(1,2,1)$ and $(0,4,0)$ is: 
$$L = OP + PQ \, t;$$
$$L = (1,2,1) + (-1,2,-1) \, t .$$
However after this I'm not sure how to proceed. I can find PA_ then draw a line from $A$ to the line $L$... advice?
 [/INST] The equation of the plane passing through the points $P$, $Q$, and $A$ is given by
$\begin{vmatrix}
x & y & z & 1 \\ 
1 & 2 & 1 & 1 \\ 
0 & 4 & 0 & 1 \\ 
8 & 3 & 2 & 1 
\end{vmatrix}=0$.
After you compute the determinant, you get $3x-6y-15z+24=0$. Then the area $S$ of the triangle defined by the points is given by $S=\frac{1}{2}\sqrt{3^2+(-6)^2+(-15)^2}=\frac{1}{2}\sqrt{270}$. The area also satisfies $S=\tfrac{1}{2}hb$, where the height $h$ is measured from $A$ to the base $b$ connecting $P$ and $Q$. So $h=2S/b$ where $b=\sqrt{(1-0)^2+(2-4)^2+(1-0)^2}=\sqrt{6}$ is the distance from $P$ to $Q$. So, you get $h=\sqrt{270}/\sqrt{6}=3\sqrt{5}$, which is the shortest distance from $A$ to the line passing through $P$ and $Q$. 
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