JEE Main 2026 Apr 5 Shift 1, Mathematics Q14: Scalar and Vector Triple Product

Rewrite the cross-product equation: $\vec{r}\times \vec{a}=\vec{b}\times \vec{a}$, equivalently $(\vec{r}-\vec{b})\times \vec{a}=\vec{0}$, so $\vec{r}=\vec{b}+\lambda \vec{a}$ for some scalar $\lambda$.

Use $\vec{r}\cdot \vec{a}=0$: $\vec{b}\cdot \vec{a}+\lambda |\vec{a}{|}^2=0$. Here $\vec{a}\cdot \vec{b}=0+1-2=-1$ and $|\vec{a}{|}^2=7+1+1=9$. So $\lambda =\frac{1}{9}$.

Then $|\vec{r}{|}^2=|\vec{b}{|}^2+{\lambda }^2|\vec{a}{|}^2+2\lambda (\vec{a}\cdot \vec{b})=5+\frac{9}{81}-\frac{2}{9}=5+\frac{1}{9}-\frac{2}{9}=\frac{44}{9}$.

Therefore $|3\vec{r}{|}^2=9\cdot \frac{44}{9}=44$.