The equation to the plane through the line of intersection of 2x+y+3z−2=0,x−y+z+4=0 such that each plane is at a distance of 2 unit from the origin is

Equation of the plane is (2x+y+3z−2)+k(x−y+z+4)=0p=|−2+4k|(2+k)2+(1−k)2+(3+k)2=2⇒k=13,−1∴ Planes are (2x+y+3z−2)+13(x−y+z+4)=0,(2x+y+3z−2)−1(x−y+z+4)=0⇒15x−12y+16z+50=0,x+2y+2z−6=0