JEE Main 2025 — Vector Algebra Question with Solution

$\begin{aligned} & \vec{a}=\hat{i}+\hat{j}+\hat{k} \\ & \vec{b}=2 \hat{i}+2 \hat{j}+\hat{k} \\ & \vec{d}=\vec{a} \times \vec{b} \\ & =-\hat{i}+\hat{j} \\ & |\vec{c}-2 \vec{a}|^2=8 \\ & |\vec{c}|^2+4|\vec{a}|^2-4 \vec{a} \cdot \vec{c}=8 \\ & |\vec{c}|^2+12-4|\vec{c}|=8 \\ & |\vec{c}|^2-4|\vec{c}|+4=0 \\ & |\vec{c}|^2=2 \\ & \vec{d}=\vec{a} \times \vec{b} \\ & \vec{d} \times \vec{c}=(\vec{a} \times \vec{b}) \times \vec{c} \\ & \left(|\vec{d}| \times|\vec{c}| \sin \frac{\pi}{4}\right)^2=((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a})^2 \\ & 4=4|\vec{b}|^2+(\vec{b} \cdot \vec{c}) 2\left(|\vec{a}|^2\right)-2(\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{b}) \end{aligned}$ Let $\vec{b}\cdot \vec{c}=x$ $\begin{aligned} & 4=36+3 x^2-20 x \\ & 3 x^2-20 x+32=0 \\ & x=\frac{8}{3}, 4 \\ & \Rightarrow \vec{b} \cdot \vec{c}=\frac{8}{3}, 4 \\ & \Rightarrow \vec{b} \cdot \vec{c}=\frac{8}{3} \end{aligned}$ Now, $|10-3\vec{b}\cdot \vec{c}|+|\vec{d}\times \vec{c}{|}^2$ $=|10-8|+(2{)}^2=6$