JEE Main 2025 — Vector Algebra Question with Solution

let ${\vec{a}}_{11}=\text{ component of }\vec{a}\text{ along }\vec{b}$ ${\vec{a}}_1=$ component of $\vec{a}$ perpendicular to $\vec{b}$ $\begin{aligned} & \overrightarrow{\mathrm{a}}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{a}}_1=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}}) \\ & \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_{11}+\overrightarrow{\mathrm{a}}_1 \end{aligned}$ $\begin{array}{rl}& \begin{array}{l}\therefore \overrightarrow{\mathrm{a}}=\\ =\frac{16}{11}(3\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4\hat{\mathrm{i}}-5\hat{\mathrm{j}}-17\hat{\mathrm{k}})\\ \overrightarrow{11}\hat{\mathrm{j}}-\frac{33}{11}\hat{\mathrm{k}}\end{array}\\ & \begin{array}{r}\alpha =4\hat{\mathrm{i}}+\hat{\mathrm{j}}-3\hat{\mathrm{k}}\end{array}\\ & {\alpha }^2+{\beta }^2+{\gamma }^2=16+1+9=26\end{array}$