JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Matrices

For the three planes to share infinitely many points the coefficient determinant must vanish. Expanding $\left|\begin{array}{ccc}1& 2& 3\\ 4& 3& -4\\ 8& 4& -\lambda \end{array}\right|=5\lambda -72$ and setting it to zero gives $\lambda =\frac{72}{5}$. Eliminating $x$ from the first pair, $4\times (1)-(2)$ yields $5y+16z=8$; from the last pair, $2\times (2)-(3)$ yields $2y+(\lambda -8)z=-1-\mu$. With $\lambda =\frac{72}{5}$ this second relation is $2y+\frac{32}{5}z=-1-\mu$, i.e. $5y+16z=\frac{5}{2}(-1-\mu )$. Consistency with $5y+16z=8$ forces $\frac{5}{2}(-1-\mu )=8$, so $\mu =-\frac{21}{5}$. The ordered pair is $\left(\frac{72}{5},-\frac{21}{5}\right)$.