Question

The equation of plane passes through the point $(-1,0,1)$ and containing the line $\frac{x+1}{2}=\frac{y-3}{-1}=\frac{z+1}{1},$ is

• A. $3x-y-6z+4=0$
• B. $x-2y-4z+7=0$
• C. $x-4y-6z+7=0$
• D. $x-3y-8z+6=0$

Answer 1

The correct option is C $x-4y-6z+7=0$

The equation of plane passes through point $(x_1,y_1,z_1)=(-1,0,1)$ is :
$a(x+1)+by+c(z-1)=0$ it containing the line $\frac{x+1}{2}=\frac{y-3}{-1}=\frac{z+1}{1}$
so, point $(x_2,y_2,z_2)=(-1,3,-1)$ satisfies the plane, and is perpendicular to $2\hat{i}-\hat{j}+\hat{k}$
So, equation of plane is :
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ a& b& c\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x+1& y-0& z-1\\ 0& 3& -2\\ 2& -1& 1\end{array}\right|=0\Rightarrow x-4y-6z+7=0$