Let \(\vec{a}=3 \hat{i}-\hat{j}+2 \hat{k}, \vec{b}=\vec{a} \times(\hat{i}-2 \hat{k})\) and \(\vec{c}=\vec{b} \times \hat…

\(\text{Given,}\\ \vec{a}=3\hat{i}-\hat{j}+2\hat{k},\\ \vec{b}=\vec{a}\times \left(\hat{i}-2\hat{k}\right)\text{ }\\ \text{and }\vec{c}=\vec{b}\times \hat{k}\\ \text{Now, }\\ \vec{\mathrm{b}}=\vec{\mathrm{a}}\times \left(\hat{\mathrm{i}}-3\hat{\mathrm{k}}\right)\\ =\left|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2\end{matrix}\right|=2\hat{i}+8\hat{j}+\hat{k}\\ \vec{\mathrm{c}}=\vec{\mathrm{b}}\times \hat{\mathrm{k}}=8\hat{\mathrm{i}}-2\hat{\mathrm{j}}\\ \vec{\mathrm{c}}-2\hat{\mathrm{j}}=8\hat{\mathrm{i}}-4\hat{\mathrm{j}}\\ \text{ Projection of  }\left(\vec{\mathrm{c}}-2\hat{j}\right)\text{ on }\vec{a}\\ \left(\vec{\mathrm{c}}-2\hat{\mathrm{j}}\right)\cdot \hat{\mathrm{a}}=\frac{⟨8,-4,0⟩\cdot ⟨3,-1,2⟩}{\sqrt{14}}\\ =\frac{28}{\sqrt{14}}=2\sqrt{14}\)