Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned} 0\\-\frac{3}{2} \\-1\\ 1 \end{aligned}\right].$$
(b) The dimension of the solution space is $1$.
Work Step by Step
The coefficient matrix is given by
$$\left[ \begin {array}{cccc} -1&2&-1&2\\ -2&2&1&4
\\ 3&2&2&5\\ -3&8&5&17\end {array}
\right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&0&0\\ 0&1&0&\frac{3}{2}
\\ 0&0&1&1\\ 0&0&0&0\end {array}
\right]
.
$$
The corresponding system is
$$
\begin{aligned}
x_1 &=0\\
x_2+\frac{3}{2}x_4 &=0\\
x_3+x_4 &=0\\
\end{aligned}.
$$
The solution of the above system is $x_1= 0$,$x_2=-\frac{3}{2}t$,$x_3=-t$, $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned} 0\\-\frac{3}{2}t\\-t\\ t \end{aligned}\right] = t \left[\begin{aligned} 0\\-\frac{3}{2} \\-1\\ 1 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned} 0\\-\frac{3}{2} \\-1\\ 1 \end{aligned}\right].$$
(b) The dimension of the solution space is $1$.