If the line $\vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ is parallel to the plane $\vec{r} . (3 \hat{i}-2 \hat{j}+a \hat{k})=11$, then a is equal to :

-2

$\vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$

this vector is parallel to direction of line

Plane: $\vec{r}-(3 \hat{i}-2 \hat{j}+a k)=11$

this vector is perpendicular to place

let $\vec{p} =2 \hat{i}+\hat{j}+2 \hat{k}$

$\vec{q} =3 \hat{i}-2 \hat{j}+a \hat{k}$

as $\vec{p}$ is parallel to plane

⇒ $\vec{p}$ is perpendicular to $\vec{q}$

⇒ $\vec{p} . \vec{q}=0$

(since angle $\vec{p}$ and $\vec{q}$ is 90° and $\vec{p}.\vec{p} = |\vec{p}||\vec{q}|cos90°=0$)

$\Rightarrow (2 \hat{i}+\hat{j}+2 \hat{k})(3 \hat{i}-2 \hat{j}+a \hat{k})=0$

$6-2+2 a=0$

$4+2 a=0$

$2 a=-4$

$a=-2$