Question

Find the distance of point $(-2,3,-4)$ from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ measured parallel to the plane $4x+12y-3z+1=0$.

Answer 1

$(-2,3,-4)\frac{x+2}{3}=\frac{2y+3}{4}=\frac{37+4}{5}$

$4x+12y-37+1=0$

Let $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{37+4}{5}=k$

$Q(3k-2,\frac{4k-3}{2},\frac{5k-4}{3})$

$4x+12y-37+1=0$

$4(3k-2)+12(4k-3)/23(5k-4)/3=0$

$k=2$

$Q(4,5/2,2)$

$PQ=17/2$