Given that,
$$
x_1=3, y_1=3 \text { and } x_2=7, y_2=6
$$
$\therefore$ Equation of straight line,
$$
\begin{aligned}
\quad\left(y-y_1\right) & =\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\
\Rightarrow \quad(y-3) & =\frac{6-3}{7-3}(x-3) \\
\Rightarrow \quad(y-3) & =\frac{3}{4}(x-3)
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & 4 y-12=3 x-9 \\
\Rightarrow & 3 x-4 y=-3
\end{array}
$$
For finding the points of intersection, put $x=0$
then
$$
\begin{aligned}
3 \times 0-4 y & =-3 \\
-4 y & =-3 \\
y & =\frac{3}{4}
\end{aligned}
$$
$$
\Rightarrow
$$
$$
\Rightarrow
$$
$$
y=\frac{3}{4}
$$
and when $y=0$
$$
\begin{array}{rlrl}
& \Rightarrow & 3 x & =-3 \\
\Rightarrow & x & =-1
\end{array}
$$
Hence, the points are $A\left(0, \frac{3}{4}\right)$ and $B(-1,0)$.
$$
\begin{aligned}
\therefore \text { Length of } A B & =\sqrt{(0+1)^2+\left(\frac{3}{4}-0\right)^2} \\
& =\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}
\end{aligned}
$$