JEE Main 2024 — 9 April, Shift 2 — Question 20

Area of parallelogram $=\frac{1}{2} | \overrightarrow{\mathrm{d}}_{1} \times \overrightarrow{\mathrm{d}}_{2}|$

$A=\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})|=\frac{\sqrt{21}}{2}$

so, $\vec{a}+\vec{b}=\hat{i}+\alpha \hat{j}+2 \hat{k}$

$\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+\beta \hat{\mathrm{j}}$

\left( \vec{a}+\vec{b} \right)\times \left( \vec{b}+\vec{c} \right)=\left| \begin{array}{*{35}{l}}\overset{}{\mathop{i}}\, & \overset{}{\mathop{j}}\, & \overset{}{\mathop{k}}\, \\1 & \alpha & 2 \\-1 & \beta & 0 \\\end{array} \right|

$|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})|=\sqrt{4 \beta^{2}+4+(\alpha+\beta)^{2}}=\sqrt{21}$

$4 \beta^{2}+4+\alpha^{2}+\beta^{2}+2 \alpha \beta=21$

$\alpha^{2}+5 \beta^{2}-12=17$

$\alpha^{2}+5 \beta^{2}=29$ and $\alpha \beta=-6$ and given $\alpha_{i} \beta$ are integers

so, $\alpha=-3, \beta=2$ or $\alpha=3, \quad \beta=-2$

$\left(\alpha_{1}, \beta_{1}\right)=(-3,2)$

$\left(\alpha_{2}, \beta_{2}\right)=(3,-2)$

$\alpha_{1}^{2}+\beta_{1}^{2}-\alpha_{2} \beta_{2}=9+4+6=19$