JEE Main 2025 — 2 April, Evening Shift — Question 39

$2 x+\lambda y+3 z=5$

$3 x+2 y-z=7$

$4 x+5 y+\mu z=9$

For infinite solutions $\Rightarrow \Delta=0=\Delta_{1}=\Delta_{2}=\Delta_{3}$

$\Delta=\left|\begin{array}{ccc}2 & \lambda & 3\\ 3 & 2 & -1\\ 4 & 5 & \cdots\end{array}\right|=0$

$\Rightarrow-4 \lambda-3 \lambda \mu+4 \mu+31=0$

$\Delta_{1}=\left|\begin{array}{ccc}5 & \lambda & 3\\ 7 & 2 & -1\\ 9 & 5 & \mu\end{array}\right|=0 \Rightarrow-9 \lambda-7 \lambda \mu+10 \mu+76=0$

$\Delta_{2}=\left|\begin{array}{ccc}2 & 3 & 5\\ 3 & -1 & 7\\ 4 & \mu & 9\end{array}\right|=0 \Rightarrow \mu+5=0 \Rightarrow \mu=-5$

$\Delta_{3}=\left|\begin{array}{lll}2 & \lambda & 5\\ 3 & 2 & 7\\ 4 & 5 & 9\end{array}\right|=0 \Rightarrow \lambda+1=0 \Rightarrow \lambda=-1$

$\therefore$ For infinite solution $\mu=-5$ and $\lambda=-1$

Now $\mu^{2}+\lambda^{2}=25+1$

$$=26$$