Answer
$$-3\left[\begin{array}{l}{1} \\ {-1}\\{0} \end{array}\right]+ 2\left[\begin{array}{l}{2} \\ {0} \\{1} \end{array}\right]+0\left[\begin{array}{l}{4} \\ {2}\\{3} \end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\\{2} \end{array}\right].$$
Work Step by Step
Assume that
$$x_1\left[\begin{array}{l}{1} \\ {-1}\\{0} \end{array}\right]+x_2\left[\begin{array}{l}{2} \\ {0} \\{1} \end{array}\right]+x_3\left[\begin{array}{l}{4} \\ {2}\\{3} \end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\\{2} \end{array}\right],$$
then we have the system
$$
\begin{aligned}
x_{1}+2x_{2}+4 x_{3} &=1 \\
-x_{1} +2x_3 &=3\\
x_{2}+3 x_{3} &=2
\end{aligned}.
$$
The augmented matrix is given by
$$
\left[ \begin {array}{cccc} 1&2&4&1\\ -1&0&2&3
\\ 0&1&3&2\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[ \begin {array}{cccc} 1&0&-2&-3\\ 0&1&3&2
\\ 0&0&0&0\end {array} \right]
.$$
From which the solution is
$$x_1=-3+2t, \quad x_2=2-3t, \quad x_3=t.$$
For example, take $t=0$, then
$$-3\left[\begin{array}{l}{1} \\ {-1}\\{0} \end{array}\right]+ 2\left[\begin{array}{l}{2} \\ {0} \\{1} \end{array}\right]+0\left[\begin{array}{l}{4} \\ {2}\\{3} \end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\\{2} \end{array}\right].$$
