If the system of equations \(\begin{aligned}& x+2 y-3 z=2 \\& 2 x+\lambda y+5 z=5 \\& 14 x+3 y+\mu z=33\end{aligned}\) h…

Given

\(x+2y-3z=2\)

\(2x+\lambda y+5z=5\)

\(14x+3y+\mu z=33\)

For infinitely many solutions, the third equation must be a linear combination of the first two:

Checking the standard condition for infinitely many solutions using determinants,

\(\Delta=\begin{vmatrix}1\&2\&-32\&\lambda\&514\&3\&\mu\end{vmatrix}=0\)

\(\Delta_x=\Delta_y=\Delta_z=0\)

\(\lambda=4,\qquad \mu=8\)

\(\lambda+\mu=12\)