JEE Main 2025 — 28 January, Morning Shift — Question 23

$\vec{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

$=-\hat{i}+\hat{j}$

$|\overrightarrow{\mathrm{c}}-2 \overrightarrow{\mathrm{a}}|^{2}=8$

$|c|^{2}+4|a|^{2}-4(a . c)=8$

$c^{2}+12-4 c=8$

$c^{2}-4 \mathrm{c}+4=0$

$|c|=2$

$\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

$\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}$

$\left(|\mathrm{d}||\mathrm{c}| \sin \frac{\pi}{4}\right)^{2}=((\mathrm{a} . \mathrm{c}) \cdot \mathrm{b}-(\mathrm{b} . \mathrm{c}) \cdot \mathrm{a})^{2}$

$4=4 b^{2}+(b.c)^2.(a)^2-2(b . c)(a . b)$

$4=36+3 \mathrm{x}^{2}-20 \mathrm{x}$

Let b.c $=\mathrm{x}$

$3 x^{2}-20 x+32=0$

$3 x^{2}-12 x-8 x+32=0$

$\mathrm{x}=\frac{8}{3}, 4$

b.c $=\frac{8}{3}, 4$

b.c $=\frac{8}{3}$

Now $|10-3 \mathrm{~b} . \mathrm{c}|+|\mathrm{d} \times \mathrm{c}|^{2}$

$|10-8|+(2)^{2}$

$\Rightarrow 6$ Ans.