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

P₁: 2x+3y+z - 4=0 and P₂: x+y+z - 2 = 0 is $P_1+\lambda P_2 = 0$ .

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

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

Now, this plane is perpendicular to plane 2x+3y−z = 3.

So, sum of product of their direction ratios is zero.

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

Put value of $\lambda = -3$ in equation (1), we get the Equation of required plane as

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