JEE Main 2024 — 31 January, Shift 2 — Question 24

$(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$

$(5 \hat{i}+\hat{j}+4 \hat{k}) \times \vec{c}=2(7 \hat{i}-7 \hat{j}-7 \hat{k})+24 \hat{j}-6 \hat{k}$

$\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} 5 & 1 & 4 x & y & z\end{array}\right|=14 \hat{i}+10 \hat{j}-20 \hat{k}$

$\Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}}$

$z-4 y=14,4 x-5 z=10,5 y-x=-20$

$(a-b+i) \cdot \vec{c}=-3$

$(2 \hat{i}+3 \hat{j}-2 \hat{k}) \cdot \vec{c}=-3$

$2 x+3 y-2 z=-3$

$\therefore \mathrm{x}=5, \mathrm{y}=-3, \mathrm{z}=2$

$|\overrightarrow{\mathrm{c}}|^{2}=25+9+4=38$