Answer
The solution of the system is $\left( 1,2,3 \right)$.
Work Step by Step
Let us consider the given equation:
$3x+2y-3z=-2$ …… (I)
$2x-5y+2z=-2$ …… (II)
$4x-3y+4z=10$ …… (III)
Calculate $2\times $ equation (II) minus equation (III):
$\begin{align}
& 4x-10y+4z=-4 \\
& 4x-3y+4z=10 \\
& -\text{ }+\text{ }-\text{ }- \\
& \overline{\,\,\,\,\,\,\,\,\,\,-7y=-14\,\,\,\,\,\,\,} \\
\end{align}$
Further simplify,
$y=2$
Apply $2\times $ equation (I) and add $3\times $ equation (II):
$\begin{align}
& 6x+4y-6z=-4 \\
& 6x-15y+6z=-6 \\
& \overline{\,\,\,\,\,\,\,\,\,12x-11y=-10\,\,\,} \\
\end{align}$
$12x-11y=-10$ …… (IV)
Putting the value of y in equation (IV):
$\begin{align}
& 12x-11(2)=-10 \\
& 12x=-10+22 \\
& 12x=12 \\
& x=1
\end{align}$
Now, putting the value $x=1,y=2$ in equation (I):
$\begin{align}
& 3\left( 1 \right)+2\left( 2 \right)-3z=-2 \\
& 3+4-3z=-2 \\
& -3z=-9 \\
& z=3
\end{align}$
Thus, the order triple $\left( 1,2,3 \right)$ satisfies the three equations.
