Let the system of equations x + 2y + 3z = 5, 2x + 3y + z = 9, 4x +...

Given: $x + 2y + 3z = 5$, $2x + 3y + z = 9$ and $4x + 3y + \lambdaz = \mu$.

For infinite solutions, $\Delta = {\Delta}_{1} = {\Delta}_{2} = {\Delta}_{3} = 0$.

⇒ \Delta = \left|\begin{cases}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{cases}\right| = 0

$⇒ 3\lambda - 3 - 4\lambda + 8 + 18 - 36 = 0$

$⇒ \lambda = - 13$

Now, finding 

⇒ {\Delta}_{1} = \left|\begin{cases}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & - 13\end{cases}\right| = 0

$⇒ \mu = 15$

Now, finding

⇒ {\Delta}_{2} = \left|\begin{cases}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & \mu & \lambda\end{cases}\right|

⇒ {\Delta}_{2} = \left|\begin{cases}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & - 13\end{cases}\right|

$⇒ {\Delta}_{2} = 0$

And 

⇒ {\Delta}_{3} = \left|\begin{cases}1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & \mu\end{cases}\right|

⇒ {\Delta}_{3} = \left|\begin{cases}1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15\end{cases}\right|

$⇒ {\Delta}_{3} = 0$

For $\lambda = - 13, \mu = 15$ system of equation has infinite solutions.

$⇒ \lambda + 2\mu = 17$.