If with reference to the right-handed system of mutually perpendicular unit vectors $\hat{i}, \hat{j}$ and $\hat{k}$,$\vec{\alpha} = 3\hat{i} - \hat{j}$ and $\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}$,then express $\vec{\beta}$ in the form $\vec{\beta} = \vec{\beta}_{1} + \vec{\beta}_{2}$,where $\vec{\beta}_{1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_{2}$ is perpendicular to $\vec{\alpha}$.

Let $\vec{\beta}_{1} = \lambda \vec{\alpha}$,where $\lambda$ is a scalar. Then $\vec{\beta}_{1} = 3\lambda \hat{i} - \lambda \hat{j}$.
Now,$\vec{\beta}_{2} = \vec{\beta} - \vec{\beta}_{1} = (2 - 3\lambda) \hat{i} + (1 + \lambda) \hat{j} - 3\hat{k}$.
Since $\vec{\beta}_{2}$ is perpendicular to $\vec{\alpha}$,we have $\vec{\alpha} \cdot \vec{\beta}_{2} = 0$.
$3(2 - 3\lambda) - 1(1 + \lambda) = 0$
$6 - 9\lambda - 1 - \lambda = 0$
$5 - 10\lambda = 0 \implies \lambda = \frac{1}{2}$.
Thus,$\vec{\beta}_{1} = \frac{3}{2}\hat{i} - \frac{1}{2}\hat{j}$ and $\vec{\beta}_{2} = \frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}$.