JEE Main 2025 — 7 April, Evening Shift — Question 39

$\Delta=\left|\begin{array}{ccc}1 & 5 & -1\\ 4 & 3 & -3\\ 24 & 1 & \lambda\end{array}\right|$

$=(3 \lambda+3)-5(4 \lambda+72)-(4-72)$

$=-17 \lambda-289=0$

$\Rightarrow \lambda=-17$

$\Delta_{1}=\left|\begin{array}{ccc}1 & 5 & -1\\ 7 & 3 & -3\\ \mu & 1 & -17\end{array}\right|$

$=(-51+3)-5(-119+3 \mu)-(7-3 \mu)$

$=540-12 \mu$ $\mu=\frac{540}{12}=45$

$\therefore x+5 y-z=1$

$4 x+3 y-3 z=7$

$24 x+y-17 z=45$

$z=x+5 y-1$

$\therefore 4 x+3 y-3 x-15 y+3=7$

$\Rightarrow x-12 y=4$

$\Rightarrow x=4+12 y$

$\therefore z=4+12 y+5 y-1$ $=3+17 y$

$\therefore(x, y, z) \equiv(4+12 k, k, 3+17 k)(\because$ Assume $y=k)$

$7 \leq 7+30 k \leq 77$

$0 \leq 30 k<70$

$0 \leq k \leq 2 \cdot 3$

$\Rightarrow k=0,1,2$ Three solutions possible