JEE Main 2025 — 28 January, Evening Shift — Question 3

let $\vec{a}_{11}=$ component of $\vec{a}$ along $\vec{b}$

$\vec{a}_{1}=$ component of $\vec{a}$ perpendicular to $\vec{b}$

$\overrightarrow{\mathrm{a}}_{11}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$

$\overrightarrow{\mathrm{a}}_{1}=\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}})$

$\because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{a}}_{11}+\overrightarrow{\mathrm{a}}_{1}$

$\therefore \overrightarrow{\mathrm{a}}=\frac{16}{11}(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\frac{1}{11}(-4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-17 \hat{\mathrm{k}})$

$=\frac{44}{11} \hat{\mathrm{i}}+\frac{11}{11} \hat{\mathrm{j}}-\frac{33}{11} \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{a}}=4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

$\alpha=4 \quad \beta=1 \quad \gamma=-3$

$\alpha^{2}+\beta^{2}+\gamma^{2}=16+1+9=26$