JEE Main 2024 — 30 January, Shift 1 — Question 13

$\mathrm{x}+\mathrm{y}+\mathrm{z}=4 \mu, \mathrm{x}+2 \mathrm{y}+2 \lambda \mathrm{z}=10 \mu, \mathrm{x}+3 \mathrm{y}+4 \lambda{ }^{2} z=\mu^{2}+15$

$\Delta=\left|\begin{array}{ccc}1 & 1 & 1\\ 1 & 2 & 2\\ \lambda 1 & 3 & 4 \lambda^{2}\end{array}\right|=(2 \lambda-1)^{2}$

For unique solution $\Delta \neq 0,2 \lambda-1 \neq 0,\left(\lambda \neq \frac{1}{2}\right)$

Let $\Delta=0, \lambda=\frac{1}{2}$

$\Delta_{\mathrm{y}}=0, \Delta_{\mathrm{x}}=\Delta_{\mathrm{z}}=\left|\begin{array}{ccc}4 \mu & 1 & 1 \\10 \mu & 2 & 1 \\\mu^{2}+15 & 3 & 1\end{array}\right|$

$=(\mu-15)(\mu-1)$

For infinite solution $\lambda=\frac{1}{2}, \mu=1$ or 15