Let the line L be the projection of the line x-12=y-31=z-42 in the plane x-2y-z=3.

Question

Let the line $L$ be the projection of the line $\dfrac{x-1}{2}=\dfrac{y-3}{1}=\dfrac{z-4}{2}$ in the plane $x-2y-z=3$. If $d$ is the distance of the point $(0,0,6)$ from $L$, then $d^2$ is equal to

Answer: 26

Accepted Answer

Foot of perpendicular from $(1,3,4)$ to the plane $x-2y-z=3$: normal $(1,-2,-1)$,
$$t=-\frac{(1-6-4)-3}{1+4+1}=-\frac{-12}{6}=2,$$
giving foot $A=(1,3,4)+2(1,-2,-1)=(3,-1,2)$.
Intersection of the given line with the plane: point $(2\lambda+1,\lambda+3,2\lambda+4)$, substituting:
$$(2\lambda+1)-2(\lambda+3)-(2\lambda+4)=3\implies -2\lambda-9=3\implies\lambda=-6,$$
giving $B=(-11,-3,-8)$.
Line $L=AB$: direction $B-A=(-14,-2,-10)\parallel(7,1,5)$, through $A(3,-1,2)$: point $C=(7\mu+3,\mu-1,5\mu+2)$.
Foot of perpendicular from $(0,0,6)$: $(C-(0,0,6))\cdot(7,1,5)=0$:
$$7(7\mu+3)+(\mu-1)+5(5\mu-4)=0\implies 75\mu=0\implies\mu=0,\ C=(3,-1,2).$$
$d^2=(3)^2+(-1)^2+(2-6)^2=9+1+16=26.$

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