QJEEM24Mac86a: JEE Main 2024 Physics - Class 11 JEEM24Mac86a Solution: Let and be a vector such that . If , then is equal to:

$\begin{aligned}& \left.\begin{array}{l}\vec{a}=4 \hat{i}-\hat{j}+\hat{k} \\\vec{b}=11 \hat{i}-\hat{j}+\hat{k}\end{array}\right] \\& \vec{a} \cdot \vec{b}=44+1+1=46 \\& |\vec{a}|^2=18,\left|\vec{b}^2\right|=123 \\& (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \\& (2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670 \\& (\vec{a}+\vec{b}) \times c=(2 \vec{a}-3 \vec{b}) \times \vec{c} \\& (-\vec{a}+4 \vec{b}) \times \vec{c}=0 \\& \vec{c}=\lambda(4 \vec{b}-\vec{a}) \\& (2 \vec{a}+3 \vec{b}) \cdot \lambda(4 \vec{b}-\vec{a})=1670 \\& \lambda\left(5 \vec{a} \cdot \vec{b}-2|\vec{a}|^2+12|\vec{b}|^2\right)=1670 \\& \lambda=\frac{1670}{5 \times 46-2 \times 18+12 \times 123} \\& \lambda=1 \\& \vec{c}=4 \vec{b}-\vec{a} \\& =4(11 \hat{i}-\hat{j}+\hat{k})-(4 \hat{i}-\hat{j}+\hat{k}) \\& =40 \hat{i}-3 \hat{j}+3 \hat{k} \\& \left|\vec{c}^2\right|=1600+9+9 \\& =1618 \\\end{aligned}$