Question

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

Answer 1

Let the equation of plane be $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

If passes through $(2,2,1)$

$\therefore$ Equation is $a(x-2)+b(y-2)+c(z-1)=0$

Vector normal to plane is the cross product of vectors

$(2\hat{i}-\hat{j}+2k)$ and $(\hat{i}+\hat{j}-\hat{k})$

$\overrightarrow{n}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 2\\ 1& 1& -1\end{array}\right]$

$=\hat{i}(1-2)-\hat{j}(-3-2)+\hat{k}(3+1)$

$=-\hat{i}+5\hat{j}+4\hat{k}$

$\therefore$ Equation of plane is $a(x-2)+b(y-2)+c(z-1)=0$

$\Rightarrow -1(x-2)+5(y-2)+4(z-1)=0$

$\Rightarrow -x+2+5y-10+4z-4=0$

$\Rightarrow -x+5y+4z-12=0$

$\Rightarrow x-5y-4z+12=0$