Sum of the distances from 3 points to a line which through a fixed point

Question

In $\mathbb R^3$, let $A(1,0,0)$, $B(0,2,0)$, $C(0,0,3)$ and $D(-2,-3,-4)$. Find the equation of the line $\Delta$, which $D\in \Delta$ and sum of distances from $A,B,C$ to $\Delta$ attains a maximum ?

I have no idea for this question. Anyone have an idea?

Answer 1

Idea:

Write the equations of of spheres with diameters $AD,BD$ and $CD$ and calculate their intersection point, name it $E$. I bet $\Delta$ is $DE$.

Hmm, I just realized that is the same as writing of equation of plane through $A,B$ and $C$ and then write the perpendicular to this plane through $E$.

Answer 2

Here's a dumb brute-force approach ...

We know the line $\Delta$ goes through $D$, so we only need to decide its direction. Suppose we denote the direction by the unit vector $U=(u,v,w)$.

Let $F_A(u,v,w)$ be the distance from point $A$ to the line $\Delta$. You can obtain an expression for $F_A(u,v,w)$ by numerous methods. For example, using the formula from here, we get $$ F_A(u,v,w) = \frac{\|(A-D) \times U\|}{\|U\|} = \sqrt{(3w-4v)^2 + (3w-4u)^2 + (3v-3u)^2} $$ Similarly, you can obtain a formula for the distance $F_B(u,v,w)$ from point $B$ to the line $\Delta$, and a formula for the distance $F_C(u,v,w)$ from point $C$ to the line $\Delta$.

We need to find the values $(u,v,w)$ that maximize $F_A(u,v,w) + F_B(u,v,w) + F_C(u,v,w)$ subject to the constraint $u^2 + v^2 + w^2 = 1$. You can do this using standard calculus methods.