Find the distance of the point `P(-1,-5,-10)` from the point of interesection of the line joining the points `A(2,-1,2)` and `B(5,3,4)` with the plane is `x-y+z=5`

Explanation

The equation of the given plane is
`x-y+z=5`…………(i)
The equation of the line AB is
`(x-2)/(5-2)=(y+1)/(3+1)=(z-2)/(4-2) rArr (x-2)/3 = (y+1)/4 =(z-2)/2= lambda` (say)………….(ii)
For some value of `lambda`, let the point `Q(3lambda+2,4lambda-1,2lambda+2)` be the pont of intersection of the plane (i) and the line (ii).
Then, `(3lambda+2),4lambda-1,2lambda+2)` be the point of intersection of the plane (i) and the line (ii).
Then, `(3lambda+2)-(4lambda-1)+(2lambda+2)=5 rArr lambda=0`.
So, the required point Q is (2,-1,2).
`therefore` PQ `=sqrt((2+1)^(2)+(-1+5)^(2)+(2+10)^(2))`
`=sqrt(3^(2)+4^(2)+12^(2))=sqrt(9+16+144)`.
`=sqrt(169)=13` units.