If $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{c}$, where $\vec{a}=-\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}$ then :

x = 3

from given values

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

$\vec{c}=\hat{i}-\vec{j}$

as $(\vec{a}+\lambda \vec{b}) \perp \vec{c} \Rightarrow (\vec{a}+\lambda \vec{b}) . \vec{c}=0$

$\Rightarrow -1+3 \lambda-2-2 \lambda+0=0$

so $-3+\lambda=0$

$\lambda=3$