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 \(…

\(\begin{matrix} & x+2y+3z=5 \\ & 2x+3y+z=9 \\ & 4x+3y+\lambda z=\mu \end{matrix}\)

for infinite following \(\Delta ={\Delta }_{1}={\Delta }_{2}={\Delta }_{3}=0\)

\(\Delta =\left|\begin{matrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda \end{matrix}\right|=0\Rightarrow \lambda =-13\)

\({\Delta }_{1}=\left|\begin{matrix}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{matrix}\right|=0\Rightarrow \mu =15\)

\({\Delta }_{2}=\left|\begin{matrix}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{matrix}\right|=0\)

\({\Delta }_{3}=\left|\begin{matrix}1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15\end{matrix}\right|=0\)

for \(\lambda =-13,\mu =15\) system of equation has infinite solution hence \(\lambda+2 \mu=17\).