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

Explanation

The equation of the given plane is
`4x+12y-3z+1=0`………….(i)

Let `P(-2,3,-4)` be the given point.
Let l be the given line whose equation is
`(x+2)/3=(2y+3)/4=(3z+4)/5=lambda`, (say). ……………….(ii)
The general point on this line is `(3lambda-2,(4lambda-3)/(2),(5lambda-4)/3)`.
For some value of `lambda`, let the point `Q(3lambda-2,(4lambda-3)/2,(5lambda-4)/3)` lie on the line (ii) such that PQ is parallel to the given plane (i).
D.r.'s of PQ are `(3lambda-2+2), ((4lambda-3)/2-1), ((5lambda-4)/(3)+4)`, i.e., `3lambda,(4lambda-9)/2, (5lambda+8)/3`.
Dr.r. of the normal to the given plane are 4,12,-3.
Now, PQ is parallel to the given plane (i).
`rArr` PQ is perpendicular to the normal to the plane (i).
`rArr (4 xx 3lambda)+12 xx (4lambda-9)/2 -3 xx (5lambda-8)/3=0`.
`rArr 12lambda+(24lambda-54)-(5lambda-8)=0 rArr 31lambda=62 rArr lambda=2`.
`therefore` coordinates of Q are `(4,5/2,2)`.
`rArr PQ =sqrt((4+2)^(2)+(5/2-3)^(2)+(2-4)^(2))=sqrt((6)^(2)+(-1/2)^(2)+6^(2))`
`rArr sqrt(72+1/4)=sqrt(289/4)=17/2` units `=8.5` units.
Hence, the required distance of the given point from the given line is 8.5 units.