Example 4. Solve the system

$$
\left\{\begin{aligned}
x_{1}-2 x_{2}+3 x_{3} & =5 \\
2 x_{1}+3 x_{2}-x_{3} & =7 \\
3 x_{1}+x_{2}+2 x_{3} & =12
\end{aligned}\right.
$$
Solution. We have (we suggest verifying this independently):

$$
\begin{aligned}
& \Delta=\left|\begin{array}{rrr}
1 & -2 & 3 \\
2 & 3 & -1 \\
3 & 1 & 2
\end{array}\right|=0, \quad \Delta_{1}=\left|\begin{array}{rrr}
5 & -2 & 3 \\
7 & 3 & -1 \\
12 & 1 & 2
\end{array}\right|=0 \\
& \Delta_{2}=\left|\begin{array}{rrr}
1 & 5 & 3 \\
2 & 7 & -1 \\
3 & 12 & 2
\end{array}\right|=0, \quad \Delta_{3}=\left|\begin{array}{rrr}
1 & -2 & 5 \\
2 & 3 & 7 \\
3 & 1 & 12
\end{array}\right|=0
\end{aligned}
$$

The system is indeterminate and consistent. We will show how to use Cramer's formulas in this case.

If we add the first equation to the second, we get the system

$$
\left\{\begin{array} { r r } 
{ x _ { 1 } - 2 x _ { 2 } + 3 x _ { 3 } = } & { 5 , } \\
{ 3 x _ { 1 } + x _ { 2 } + 2 x _ { 3 } = } & { 1 2 , } \\
{ 3 x _ { 1 } + x _ { 2 } + 2 x _ { 3 } = } & { 1 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{rr}
x_{1}-2 x_{2}+3 x_{3}= & 5 \\
3 x_{1}+x_{2}+2 x_{3}= & 12
\end{array}\right.\right.
$$

Without resorting to the Jordan-Gauss method, we rewrite the system as follows (this will be the modified system):

$$
\left\{\begin{aligned}
x_{1}-2 x_{2} & =5-3 x_{3} \\
3 x_{1}+x_{2} & =12-2 x_{3} .
\end{aligned}\right.
$$

We arbitrarily declare \( x_{3} \) to be a free variable, and \( x_{1} \) and \( x_{2} \) to be basic variables. This is possible because

$$
\Delta=\left|\begin{array}{rr}
1 & -2 \\
3 & 1
\end{array}\right|=7 \neq 0
$$

Next,

$$
\begin{aligned}
& \Delta_{1}=\left|\begin{array}{rr}
5-3 x_{3} & -2 \\
12-2 x_{3} & 1
\end{array}\right|=5-3 x_{3}+24-4 x_{3}=29-7 x_{3} \\
& \Delta_{2}=\left|\begin{array}{cc}
1 & 5-3 x_{3} \\
3 & 12-2 x_{3}
\end{array}\right|=12-2 x_{3}-15+9 x_{3}=-3+7 x_{3} .
\end{aligned}
$$

Therefore, \( x_{1}=\frac{29-7 x_{3}}{7}, \quad x_{2}=\frac{-3+7 x_{3}}{7} \). The system has an infinite number of solutions.

The general solution is \( x_{1}=\frac{29}{7}-\alpha, x_{2}=-\frac{3}{7}+\alpha, x_{3}=\alpha \), or

$$
X_{0}=\left(\frac{29}{7}-\alpha,-\frac{3}{7}+\alpha, \alpha\right), \quad X_{6}=\left(\frac{29}{7},-\frac{3}{7}, 0\right)
$$