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 …

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,

\(
\Delta=
\begin{vmatrix}
1&2&-3\\
2&\lambda&5\\
14&3&\mu
\end{vmatrix}
=0
\)

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

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

\(
\lambda+\mu=12
\)