Find the distance of the point `(-1,-5,-10)` from the point of the intersection of the line `vecr = 2hati-2hatk+lambda(3hati+4hatj+2hatk)` and the plane `vecr.(hati-hatj+hatk) = 5`.

Answer

13 units.

Explanation

The cartesian equation of the planes is `(x-2)/3=(y+1)/4=(z-2)/2=lambda` (say).
A general point on line (i) is `N(3lambda+2,4lambda-1,2lambda+2)`.
The Cartesian equation of the given plan is `x-y+z=5`.
If this point lies on the given plane, we have
`(3lambda+2)-(4lambda-1)+(2lambda+2)=5 rArr lambda=0`.
`therefore` point P is `P(2,-1,2)` and the other point is `Q(-1,-5,-10)`.
`therefore` PQ =13 units.