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?

The shortest distance should intuitively be the length of the segment which connects our point to our line and forms a right angle. Consider that we can determine this first by determining a vector from a point on our line to $A$ followed by finding its projection onto $\vec{PQ}$; the orthogonal part is then just the difference between this vector and our original vector from our line to our point and its magnitude yields the length of said segment:$$d=\left\|\vec{AP}-\frac{\vec{AP}\cdot\vec{PQ}}{\|\vec{PQ}\|^2}\vec{PQ}\right\|$$

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 $Q$.