QJEEM24Mafaef: JEE Main 2024 Physics - Class 11 JEEM24Mafaef Solution: Let and be a vector such that and . Then is equal to ________.

$\begin{aligned}& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\& (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{c}}=2(7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\& \left|\begin{array}{lrr}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\5 & 1 & 4 \\x & y & \mathrm{z}\end{array}\right|=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\& \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\& \mathrm{z}-4 \mathrm{y}=14,4 \mathrm{x}-5 \mathrm{z}=10,5 \mathrm{y}-\mathrm{x}=-20 \\& (\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \overrightarrow{\mathrm{c}}=-3 \\& (2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \overrightarrow{\mathrm{c}}=-3 \\& 2 x+3 y-2 z=-3 \\& \therefore x=5, y=-3, z=2 \\& |\overrightarrow{\mathrm{c}}|^2=25+9+4=38\end{aligned}$