JEE Main 2026 — 21 January, Evening Shift — Question 15

$\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}$

$\cos (\pi-\theta)=\frac{|\overrightarrow{\mathrm{p}}|^{2}+|\overrightarrow{\mathrm{q}}|^{2}-|\overrightarrow{\mathrm{r}}|^{2}}{2|\overrightarrow{\mathrm{p}}||\overrightarrow{\mathrm{q}}|}$

$\frac{-1}{\sqrt{3}}=\frac{12+4-|\overrightarrow{\mathrm{r}}|^{2}}{2 \cdot 2 \sqrt{3} \cdot 2}$

$|\overrightarrow{\mathrm{r}}|^{2}=24$

$\therefore|\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-3 \overrightarrow{\mathrm{r}})|^{2}+3|\overrightarrow{\mathrm{r}}|^{2}$

$=|\vec{p} \times(\vec{q}-3 \vec{p}-3 \vec{q})|^{2}+72$

$=|\vec{p} \times(-3 \vec{p}-2 \vec{q})|^{2}+72$

$=|-2 \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|^{2}+72$

$=4|\overrightarrow{\mathrm{p}}|^{2}|\overrightarrow{\mathrm{q}}|^{2} \times \sin ^{2} \theta+72$

$=4 \cdot 12.4 \cdot \frac{2}{3}+72$ $=200$