Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}-\hat{j}+3 \hat{k}$
Then, $\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \cdot \vec{b}$
$\begin{aligned}
& \vec{x}=\left(\frac{(2 \hat{i}+\hat{y}-\hat{k}) \cdot(\hat{i}-\hat{j}+3 \hat{k})}{\left(\sqrt{\left((1)^2+(-1)^2+(3)^2\right)}\right)^2}\right) \cdot(\vec{i}-\hat{j}+3 \hat{k}) \\
& \vec{x}=\left(\frac{2-1+3}{(1+1+9)}\right)(\vec{i}-\hat{j}+3 \hat{k}) \\
& \vec{x}=\frac{-2}{11}(\vec{i}-\hat{j}+3 \hat{k}) \\
& \vec{y}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \cdot \vec{a} \cdot \\
& \vec{y}=\left(\frac{(2 \hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+3 \hat{k})}{\left((2)^2+(1)^2+(-1)^2\right)}\right) \cdot(2 \hat{i}+\hat{j}-\hat{k}) \\
& \vec{y}=\left(\frac{2-1-3}{4+1+1}\right) \cdot(2 \hat{i}+\hat{j}-\hat{k}) \\
& \vec{y}=\frac{-2}{6}(2 \hat{i}+\hat{j}-\hat{k})=\frac{-1}{3}(2 \hat{i}+\hat{j}-\hat{k}) .
\end{aligned}$
Angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is $\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}|}$.
$\cos \theta=\frac{(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})}{\sqrt{6} \sqrt{11}}=\frac{2-1-3}{\sqrt{66}}=\frac{-2}{\sqrt{66}}$
Take square both sides
$\cos ^2 \theta=\frac{4}{66}$
Now, $\mathrm{x}^2+\mathrm{y}^2=\frac{4}{11}(1+1+9)+\frac{1}{9}(4+1+1)$
$\begin{aligned}
& x^2+y^2=\frac{34 \times 2}{33 \times 2}=\frac{68}{66} \\
& x^2+y^2=17 \times \frac{4}{66}=17 \cos ^2 \theta \quad\{\text { from (i) }\}
\end{aligned}$