JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Vector Algebra

Coplanarity requires the scalar triple product to vanish, and the determinant with rows $(a+1,a,a)$, $(a,a+1,a)$ and $(a,a,a+1)$ equals $3a+1$, so $a=-\frac{1}{3}$. With this value $\vec{p}\cdot \vec{q}=-\frac{1}{3}$, while $\vec{r}\times \vec{q}=\frac{1}{9}(-3\hat{i}-3\hat{j}-3\hat{k})$, so $|\vec{r}\times \vec{q}{|}^2=\frac{27}{81}=\frac{1}{3}$. Substituting into $3(\vec{p}\cdot \vec{q}{)}^2-\lambda |\vec{r}\times \vec{q}{|}^2=0$ gives $3\cdot \frac{1}{9}-\lambda \cdot \frac{1}{3}=0$, that is $\frac{1}{3}-\frac{\lambda }{3}=0$, so $\lambda =1$.