Question

Find the equation of the plane through the intersection of the planes 3x-y+2z-4=0 and x+y+z-2=0 and the point(2,2,1).

Answer 1

The equation of any plane through the intersection of the planes,

3x-y+2z-4=0 and x+y+z-2=0, is

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

The plane passes through the point (2,2,1). Therefore, this point will satisfy Eq.(i).

$\therefore 3\times 2-2+2\times 1-4)+\lambda (2+2+1-2)=0\Rightarrow (6-4)+3\lambda =0\Rightarrow 2+3\lambda =0\Rightarrow \lambda =\frac{-2}{3}$

Substituting this value of $\lambda$ in Eq.(i), we obtain required plane as

$(3x-y+2z-4)-\frac{2}{3}(x+y+z-2)=0\Rightarrow 9x-3y+6z-12-2x-2y-2z+4=0\Rightarrow 7x-5y+4z-8=0$

This is the required equation of the plane.