Solve this following

Solution:

$\vec{\alpha}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$

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

$\vec{\beta}=\vec{\beta}_{1}-\vec{\beta}_{2}$

$\vec{\beta}_{1}=\lambda(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}), \vec{\beta}_{2}=\lambda(3 \hat{\mathrm{i}}+\hat{\mathrm{j}})-2 \hat{\mathrm{i}}+\mathrm{j}-3 \hat{\mathrm{k}}$

$\vec{\beta}_{2} \cdot \vec{\alpha}=0$

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

$9 \lambda-6+\lambda+1=0$

$\lambda=\frac{1}{2}$

$\Rightarrow \vec{\beta}_{1}=\frac{3}{2} \hat{i}+\frac{1}{2} \hat{j}$

$\Rightarrow \vec{\beta}_{2}=-\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{2} \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

Now $\vec{\beta}_{1} \times \vec{\beta}_{2}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3\end{array}\right|$

$=\hat{\mathrm{i}}\left(-\frac{3}{2}-0\right)-\hat{\mathrm{j}}\left(-\frac{9}{2}-0\right)+\hat{\mathrm{k}}\left(\frac{9}{4}+\frac{1}{4}\right)$

$=-\frac{3}{2} \hat{\mathrm{i}}+\frac{9}{2} \hat{\mathrm{j}}+\frac{5}{2} \hat{\mathrm{k}}$

$=\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})$

$\vec{\beta}=\vec{\beta}_{1}-\vec{\beta}_{2} \Rightarrow \vec{\beta} \cdot \hat{\alpha}=\vec{\beta}_{1} \cdot \hat{\alpha}=\left|\vec{\beta}_{1}\right|$

$\Rightarrow \vec{\beta}_{1}=(\vec{\beta} \cdot \hat{\alpha}) \hat{\alpha}$

$\Rightarrow \vec{\beta}_{2}=(\vec{\beta} \cdot \hat{\alpha}) \hat{\alpha}-\vec{\beta}$

$\Rightarrow \vec{\beta}_{1} \times \vec{\beta}_{2}=-(\vec{\beta} \cdot \hat{\alpha}) \hat{\alpha} \times \vec{\beta}$

$=\frac{-5}{10}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}) \times(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$=\frac{1}{2}(-3 \hat{i}+9 \hat{j}+5 \hat{k})$