Given, $2 x+3 y+z=5$,
$\begin{aligned}& 3 x+y+5 z=7, \\& x+4 y-2 z=3\end{aligned}$
The given system can be written as $A X=B$, where
$A=\left[\begin{array}{ccc}2 & 3 & 1 \\ 3 & 1 & 5 \\ 1 & 4 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right], B=\left[\begin{array}{l}5 \\ 7 \\ 3\end{array}\right]$
$|A|=\left|\begin{array}{ccc}2 & 3 & 1 \\ 3 & 1 & 5 \\ 1 & 4 & -2\end{array}\right|$
$\begin{aligned}& =2(-2-20)-3(-6-5)+1(12-1) \\
& =2(-22)-3(-11)+1(11) \\
& =-44+33+11=0 \\
& |A|=0
\end{aligned}$
As $|A|=0$, this system can have no solution (or) infinitely many solutions. So we check $(\operatorname{adj} A) B$.
$\operatorname{adj} A=\left[\begin{array}{ccc}-22 & 10 & 14 \\ 11 & -5 & -7 \\ 11 & -5 & -7\end{array}\right]$
$(\operatorname{adj} A) B=\left[\begin{array}{ccc}-22 & 10 & 14 \\ 11 & -5 & -7 \\ 11 & -5 & -7\end{array}\right]\left[\begin{array}{l}5 \\ 7 \\ 3\end{array}\right]$
$=\left[\begin{array}{c}
-110+70+42 \\
55-35-21 \\
55-35-21
\end{array}\right]=\left[\begin{array}{r}
2 \\
-1 \\
-1
\end{array}\right] \neq 0$
Hence, there exist no solution