If vectors $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j}$ are such that $(\vec{b} + \lambda \vec{c})$ is perpendicular to $\vec{a}$, then find the value of $\lambda$.

$-\frac{5}{8}$

The correct answer is Option (4) → $-\frac{5}{8}$ ##

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

Since $(\vec{b} + \lambda \vec{c}) \cdot \vec{a} = 0$,

$2(-1 + 3\lambda) + 2(2 + \lambda) + 3 = 0$

$-2 + 6\lambda + 4 + 2\lambda + 3 = 0$

$8\lambda + 5 = 0$

$\Rightarrow \lambda = -\frac{5}{8}$