# The equation of the plane passing through the point $(2 , 1 , -1 )$ and the line of intersection of the planes $\overrightarrow{r}. (\overrightarrow{i}+\overrightarrow{3j}-\overrightarrow{k})=0$ and $\overrightarrow{r}. (\overrightarrow{j}+\overrightarrow{2k})=0$ is

$\begin{array}{1 1}(1) x+4y-z=0 &(2)x+9y+11z=0\\(3)2x+y-z+5=0&(4)2x-y+z=0\end{array}$

The vector equations of the give planes are
$\overrightarrow{r}. (\overrightarrow{i}+\overrightarrow{3j}-\overrightarrow{k})=0$
$\overrightarrow{r}. (\overrightarrow{j}+2\overrightarrow{k})=0$
Their cartesian equations are
$(x \overrightarrow{i}+y \overrightarrow{j}+z \overrightarrow{k}).(\overrightarrow{i}+3\overrightarrow{j}- \overrightarrow{k})=0$.
$(x \overrightarrow{i}+y \overrightarrow{j}+z \overrightarrow{k}).(\overrightarrow{j}+2 \overrightarrow{k})=0$.
$x+3y-z=0$ and $y+2z=0$
The equation of the plane passing through the line of intersection of the planes.
$x+3y-z=0$ and $y+2z=0$
$x+3y-z+ \lambda(y+2z)=0$
This passes through the point (2,1,-1)
$2+3+1+ \lambda(1-2)=0$-----(1)
$- \lambda=-6 => \lambda =6$
(1) => $x +3y-z+6(y+2z)=0$
$x+9y+11z=0$
Hence 2 is the correct answer.