If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$ and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$,find $\lambda$ such that $\vec{a}$ is perpendicular to $\lambda\vec{b} + \vec{c}$.

(D) Given vectors are $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$.
First,calculate the vector $\vec{v} = \lambda\vec{b} + \vec{c}$:
$\vec{v} = \lambda(\hat{i} + \hat{j} - 2\hat{k}) + (\hat{i} + 3\hat{j} - \hat{k})$
$= (\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - (2\lambda + 1)\hat{k}$.
Since $\vec{a}$ is perpendicular to $\vec{v}$,their dot product must be zero,i.e.,$\vec{a} \cdot \vec{v} = 0$.
$(2\hat{i} - \hat{j} + \hat{k}) \cdot ((\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - (2\lambda + 1)\hat{k}) = 0$
$2(\lambda + 1) - 1(\lambda + 3) + 1(-2\lambda - 1) = 0$
$2\lambda + 2 - \lambda - 3 - 2\lambda - 1 = 0$
$-\lambda - 2 = 0$
$\lambda = -2$.