The value of $\lambda$, so that the vector $\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$ are perpendicular to each other, is :

$\frac{5}{2}$

$\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}$

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

for $\vec{a}$ to be perpendicular to $\vec{b}$

$\vec{a} . \vec{b}=0$ 

(since after between them is 90° $\vec{a} . \vec{b}=|\vec{a}||\vec{b}| \cos 90° = 0$)

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

$2-2 \lambda+3=0$

$5=2 \lambda$

$\lambda = \frac{5}{2}$