JEE Main 2025 Apr 3 Shift 2, Mathematics Q24: Scalar and Vector Triple Product

From $\vec{b}\times \vec{d}=\vec{c}\times \vec{d}$ we get $(\vec{b}-\vec{c})\times \vec{d}=\vec{0}$, so $\vec{d}$ is parallel to $\vec{b}-\vec{c}$.

$\vec{b}-\vec{c}=(3-2)\hat{i}+(-3+1)\hat{j}+(3-2)\hat{k}=\hat{i}-2\hat{j}+\hat{k}$.

Write $\vec{d}=t(\hat{i}-2\hat{j}+\hat{k})$. Apply $\vec{a}\cdot \vec{d}=4$:

$t(1\cdot 1+2\cdot (-2)+1\cdot 1)=4\Rightarrow t(-2)=4\Rightarrow t=-2$.

So $\vec{d}=-2\hat{i}+4\hat{j}-2\hat{k}$ and $|\vec{d}{|}^2=4+16+4=24$.

$|\vec{a}{|}^2=1+4+1=6$.

Use the Lagrange identity: $|\vec{a}\times \vec{d}{|}^2=|\vec{a}{|}^2|\vec{d}{|}^2-(\vec{a}\cdot \vec{d}{)}^2=6\cdot 24-16=144-16=128$.