Question

Find the equation of the plane passing through the line of intersection of the planes
2x+3y+z = 4 and x+y+z = 2, and which is perpendicular to the plane
2x+3y−z = 3.

Answer 1

The equation of the plane passing through the line of intersection of the planes $P_1\ :2x+3y+z = 4 \ and\ P_2:\ x+y+z = 2$ as

$2x+3y+z-4+\lambda(x+y+z-2)=0\\(\lambda+2 )x+(\lambda+3)y+(\lambda+1)z-2\lambda-4=0$

As this plane is perpendicular to plane $2x+3y−z = 3$ then dot product of their direction ratios must be zero

$(\lambda+2).2+(\lambda+3).3+(\lambda+1).(-1)=0\\2\lambda+4+3\lambda+9-\lambda-1=0\\4\lambda=-12\\\lambda=-3$

Substituting the value of $\lambda$ int he above equation of plane we get

$-x-2z+2=0\\x+2z-2=0$