#### Answer

This system has infinitely many solutions and all of them are given by:
$$x_1 = \frac{1}{2}(5-t),\quad x_2=4t-1;\quad x_3 = t$$
where $t$ is any real number.

#### Work Step by Step

We will transform this system into the row-echelon form performing the following steps:
Step 1: Substitute the last two equations and rewrite all of the equations reordering the terms by putting $x_1$ in the second place and $x_2$ in the first place:
\begin{align*}
x_2 + 2x_1-3x_3=&4\\
3x_2-2x_1-13x_3 = &-8\\
4x_1+2x_3=&10
\end{align*}
Step 2: Add to the second equation the first one multiplied by $-3$ to eliminate $x_2$ from the second equation:
The second equation will be
$$3x_2 - 3x_2 - 2x_1-6x_1-13x_3+9x_3 = -8-3\times 4$$
which is rewritten as
$$-8x_1-4x_3=-20$$
we can simplify this by dividing it by $-4$:
$$2x_1+x_3=5$$
Rewriting all of the equations we are left at:
\begin{align}
x_2+2x_1-3x_3 =& 4\\
2x_1+x_3 =&5\\
4x_1+2x_3=&10
\end{align}
Step 3: Divide the last equation by $2$:
\begin{align}
x_2+2x_1-3x_3 =& 4\\
2x_1+x_3 =&5\\
2x_1+x_3=&5
\end{align}
Step 4: We see that the last two equations are exactly identical. This means that we actually have $2$ conditions with $3$ unknowns. This means that we can equate one of them with parameter $t$, lets say $x_3=t$
\begin{align}
x_2+2x_1-3x_3 =& 4\\
2x_1+x_3 =&5\\
x_3= &t
\end{align}
Step 5: We can express both $x_1$ and $x_2$ in terms of $t$. From example substitute $x_3$ for $t$ in the second equation to obtain
$$x_1 = \frac{1}{2}(5-t).$$
Now substituting both $x_1$ and $x_3$ in the second equation we express $x_2$ in terms of $t$:
$$x_2 = 4t-1$$
Step 6: Just summarizing all we did we have that this system has infinitely many solutions given by
$$x_1 = \frac{1}{2}(5-t),\quad x_2=4t-1;\quad x_3 = t$$
where $t$ is any real number.