I assume you know how to solve a system of two linear equations such as
$$\left\{
\begin{array}{c}
-5x-5y=-5 \\
x+4y=-2%
\end{array}%
\right.\qquad(\ast)$$
but not how to represent it in terms of a matrix. I start with your system of 3 equations
$$\left\{
\begin{array}{c}
x-2y+3z=7 \\
2x+y+z=4 \\
-3x+2y-2z=-10.%
\end{array}%
\begin{array}{c}
\text{(eq. 1)} \\
\text{(eq. 2)} \\
\text{(eq. 3)}%
\end{array}%
\right.\qquad(\ast\ast)$$
In order to eliminate one of the variables $x,y,z$ you can replace one of the 3 equations as follows. Multiply another equation by an adequate multiplier and add the result termwise to it. Suppose you want to eliminate $z$. If you multiply eq. 2 by $2$ you obtain the equivalent equation
$$4x+2y+2z=8.$$
Now add it to eq. 3 to get the equation
$$\left( 4-3\right) x+\left( 2+2\right) y+\left( 2-2\right)
z=8-10\Leftrightarrow x+4y=-2,$$
which means you did eliminate the variable $z$. The multiplier $m=2$ was chosen so that $1$ (coeff. of $z$ in eq. 2) $\times m-2$ (coeff.of $z$ in eq.3)$=0$, i.e. $m=2$. Thus you can replace your system by the equivalent one
$$\left\{
\begin{array}{c}
x-2y+3z=7 \\
2x+y+z=4 \\
x+4y=-2%
\end{array}%
\begin{array}{c}
\text{(eq. 1)} \\
\text{(eq. 2)} \\
\text{(new eq. 3)}%
\end{array}%
\right.$$
Similarly you get
$$\left\{
\begin{array}{c}
\left( 1-6\right) x-\left( 2+3\right) y+\left( 3-3\right) z=7-12 \\
2x+y+z=4 \\
x+4y=-2%
\end{array}%
\begin{array}{c}
\text{(new eq. 1)} \\
\text{(eq. 2)} \\
\text{(new eq. 3)}%
\end{array}%
\right.$$
by multiplying eq. 2 by the multiplier $m=-3$ (so that $1\times c+3=0$, as above) and adding it to eq.1, which simplifies to
$$\left\{
\begin{array}{c}
-5x-5y=-5 \\
2x+y+z=4 \\
x+4y=-2%
\end{array}%
\begin{array}{c}
\text{(new eq. 1)} \\
\text{(eq. 2)} \\
\text{(new eq. 3)}%
\end{array}%
\right.$$
From these new eqs. 1 and 3, you can compute $x$ and $y$ (the initial system $(\ast)$ of 2 eqs. can be solved by this method too. Do you see how?). Inserting them in eq. 2 we obtain $z=4-2x-y$.