JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Vector Algebra

Here $|\vec{a}|=\sqrt{6}$, and a right-angle rotation of $\vec{a}$ in the plane of $\vec{a}$ and $\hat{j}$ gives a vector $\vec{b}$ perpendicular to $\vec{a}$ with the same length $\sqrt{6}$. The component of $\hat{j}$ perpendicular to $\vec{a}$ is $\hat{j}-\frac{\vec{a}\cdot \hat{j}}{|\vec{a}{|}^2}\vec{a}=\frac{1}{3}(\hat{i}+\hat{j}-\hat{k})$, so $\vec{b}$ lies along $\hat{i}+\hat{j}-\hat{k}$; scaling to length $\sqrt{6}$ gives $\vec{b}=\sqrt{2}(\hat{i}+\hat{j}-\hat{k})$. Then $3\vec{a}+\sqrt{2}\vec{b}=-\hat{i}+8\hat{j}+\hat{k}$, and its projection on $\vec{c}=5\hat{i}+4\hat{j}+3\hat{k}$ is $\frac{(-1)(5)+(8)(4)+(1)(3)}{\sqrt{25+16+9}}=\frac{30}{5\sqrt{2}}=3\sqrt{2}$.