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Let the system of equations $x+2 y+3 z=5$, $2 x+3 y+z=9,4 x+3 y+\lambda z=\mu$ have infinite number of solutions. Then $\lambda+2 \mu$ is equal to :
$28$
$17$
$22$
$15$
Solution
$ x+2 y+3 z=5 $
$2 x+3 y+z=9 $
$ 4 x+3 y+\lambda z=\mu$
for infinite following $\Delta=\Delta_1=\Delta_2=\Delta_3=0$
$\Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13$
$\Delta_1=\left|\begin{array}{ccc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15$
$\Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0$
$\Delta_3=\left|\begin{array}{ccc}1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15\end{array}\right|=0$
for $\lambda=-13, \mu=15$ system of equation has infinite solution hence $\lambda+2 \mu=17$
