Answer
(a) A basis for the solution space is
$$ \left[\begin{aligned} 3\\-\frac{1}{2}\\-4\\ 1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.
Work Step by Step
The coefficient matrix is given by
$$
\left[ \begin {array}{cccc} 3&8&2&3\\ 4&6&2&-1
\\ 3&4&1&-3\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&0&-3\\ 0&1&0&\frac{1}{2}
\\ 0&0&1&4\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned}
x_1 - 3x_4 &=0\\
x_2+\frac{1}{2}x_4 &=0\\
x_3+4x_4 &=0\\
\end{aligned}.
$$
The solution of the above system is $x_1= 3t$,$x_2=-\frac{1}{2}t$,$x_3=-4t$, $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the form
$$x= \left[\begin{aligned} x_1\\ x_2\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned} 3t\\-\frac{1}{2}t\\-4t\\ t \end{aligned}\right] = t\left[\begin{aligned} 3\\-\frac{1}{2}\\-4\\ 1 \end{aligned}\right] .$$
(a) A basis for the solution space is
$$ \left[\begin{aligned} 3\\-\frac{1}{2}\\-4\\ 1 \end{aligned}\right] .$$
(b) The dimension of the solution space is $1$.