Question

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

Answer 1

Equation of plane passing through line of intersection of given planes is given by,
$(3x-y+2z-4)+\alpha (x+y+z-2)=0$, where $\alpha \in R........\left(1\right)$

The plane passes through the point $(2,2,1)$.

Therefore, this point will satisfy equation (1),

$\therefore$ $(3\times 2-2+2\times 1-4)+\alpha (2+2+1-2)=0$

$\Rightarrow$ $2+3\alpha =0$

$\Rightarrow$ $\alpha =-\frac{2}{3}$

Substituting $\alpha =-\frac{2}{3}$ in equation (1), we obtain

$\Rightarrow$ $(3x-y+2z-4)-\frac{2}{3}(x+y+z-2)=0$

$\Rightarrow$ $(9x-3y+6z-12)-2(x+y+z-2)=0$

$\Rightarrow$ $7x-5y+4z-8=0$

This is the required equation of the plane.