$M=$ Mid point of $A C=\left(\frac{-3+h}{2}, \frac{1+k}{2}\right)$
$\because(h, k)$ lies on $7 x-4 y-1=0$.
$\Rightarrow 7 h-4 k-1=0 \Rightarrow h=\frac{4 k+1}{7}$
$$
\therefore M=\left(\frac{-3+\frac{4 k+1}{7}}{2}, \frac{1+k}{2}\right)=\left(-10+2 k, \frac{1+k}{2}\right)
$$
$\because M$ lies on $B M$.
$$
\Rightarrow 2(-10+2 k)+\left(\frac{1+k}{2}\right)-3=0 \Rightarrow k=5
$$
from equation $(2) \Rightarrow h=\frac{4 \times 5+1}{7}=3$
$$
\therefore(h, k)=(3,5)
$$
Slope of line $A C=m_1=\frac{5-1}{3+3}=\frac{4}{6}=\frac{2}{3}$
Slope of line $\mathrm{CN}=m_2=\frac{7}{4}$.
Let slope of $B C$ is $m$.
Since $\angle B C N=\angle N C A$.
$$
\Rightarrow\left|\frac{m-\frac{7}{4}}{1+\frac{7}{4} m}\right|=\left|\frac{\frac{7}{4}-\frac{2}{3}}{1+\frac{7}{4} \times \frac{2}{3}}\right|=\frac{1}{2}
$$
$$
\Rightarrow \frac{m-\frac{7}{4}}{1+\frac{7}{4} m}=\frac{1}{2} \Rightarrow m=18
$$
Equation of $B C$ is
$$
\begin{aligned}
& (y-5)=m(x-3) \Rightarrow(y-5)=18(x-3) \\
& \Rightarrow 18 x-y=49 .
\end{aligned}
$$