# The values of $\lambda$ and $\mu$ for which the equations $x+y+z=3,x+3y+2z=6,x+\lambda y+3z=\mu$ have

$\begin{array}{1 1}(A)\;\text{a unique solution if }\lambda=5\;\mu \in R&(B)\;\text{no solution if }\lambda\neq 5,\mu=9\\(C)\;\text{Infinite many solution if }\lambda=5,\mu\neq 9&(D)\;\text{None of the above}\end{array}$

Given :
$x+y+z=3$
$x+3y+2z=6$
$x+\lambda+3z=\mu$
$\Delta=\begin{vmatrix}1&1&1\\1&3&2\\1&\lambda&3\end{vmatrix}=5-\lambda$
$\Rightarrow$ If $\Delta\neq 0$ (i.e) $\lambda\neq 0$ the solution is unique.
Case-I:If $\lambda\neq 5$ and $\mu$ any real number gives unique solution.
Case-II:If $\lambda=5\Rightarrow \Delta =0$
$\Delta_1=\begin{vmatrix}3&1&1\\6&3&2\\\mu&5&3\end{vmatrix}=-\mu+9$
$\Delta_2=\begin{vmatrix}1&3&1\\1&6&2\\1&\mu&3\end{vmatrix}=-\mu +9$
$\Delta_3=\begin{vmatrix}1&1&3\\1&6&2\\1&\mu&3\end{vmatrix}=2(\mu-9)$
If $\Delta\neq 0,\mu\neq 9$.The system has no solution.
Case III:If $\lambda=5$ and $\mu=9$ equation has infinite solution.
Hence (D) is the correct answer.