Let $\overrightarrow{\mathrm{a}}=3 \hat{i}-\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times(\hat{i}-2 \hat{k})$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{k}$. Then the projection of $\overrightarrow{\mathrm{c}}-2 \hat{j}$ on $\vec{a}$ is :

<p>$$\begin{aligned} &\begin{aligned} & \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}-3 \hat{\mathrm{k}}) \\ & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{array}\right|=2 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{k}}=8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}=8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}} \end{aligned}\\ &\text { Projection of }(\hat{i}-2 \hat{j}) \text { on } \vec{a}\\ &\begin{aligned} & (\overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}) \cdot \hat{\mathrm{a}}=\frac{\langle 8,-4,0\rangle \cdot\langle 3,-1,2\rangle}{\sqrt{14}} \\ & =\frac{28}{\sqrt{14}}=2 \sqrt{14} \end{aligned} \end{aligned}$$</p>