Question:

The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x-y+z=3 and at a distance 2√3 from the point (3, 1, -1) is

x-√2y=1-√2

√2x+y=3√2-1

x+y+z=√3

5x-y+z=17

Equation of the required plane is L=(x+2y+3x-2)+K(x-y+z-3)=0 ⇒(1+K)x+(2-K)y+(3+K)z-(2+3K)=0 Its distance from (3,1,-1) is 2√3