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

The correct answer is $\overrightarrow{r} . (2\hat{i} - 2 \hat{j} + \hat{k}) =1$

The equation of plane passing through the intersection of the planes $3x-y+2z-4=0$ and $x+y+z-3=0$ cab be written as follows

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


As the plane passes through the point $(2,2,1)$, putting this value in the above equation of plane, we get

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

$6-2+2-4$ + $\lambda (2)$ = $0$

$2 + 2 \lambda$ = $0$

⇒ $\lambda = -1$


The equation of plane

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

⇒ $3x-x-y-y+2z-z-4+3=0$

⇒ $2x-2y+z=1$


The above equation in vector form

$\overrightarrow{r} . \hat{n} = 0$

$\overrightarrow{r} . (2\hat{i} - 2 \hat{j} + \hat{k}) =1$