JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Matrices

Since $A{A}^{\top }=I_3$, the rows of $A$ are orthonormal. The first row gives $4q^2+r^2=1$, and the second and third both give $p^2+q^2+r^2=1$. Orthogonality of the second and third rows requires $2q^2-r^2=0$, so $r^2=2q^2$. Substituting into $4q^2+r^2=1$ gives $6q^2=1$, hence $q^2=\frac{1}{6}$ and $r^2=\frac{1}{3}$, and then $p^2=1-\frac{1}{6}-\frac{1}{3}=\frac{1}{2}$, so $|p|=\frac{1}{\sqrt{2}}$.