$\frac{1}{2}\left|\begin{array}{lll}
0 & 0 & 1 \\
1 & 3 & 1 \\
x & y & 1
\end{array}\right|=0$

This gives      $\frac{1}{2}(y-3 x)=0 \text { or } y=3 x$

which is the equation of required line $\mathrm{AB}$.

Also, since the area of the triangle $\mathrm{ABD}$ is $3$ sq. units, we have

$\frac{1}{2}\left|\begin{array}{lll}
1 & 3 & 1 \\
0 & 0 & 1 \\
k & 0 & 1
\end{array}\right|=\pm 3$

This gives, $\frac{-3 k}{2}=\pm 3,$ i.c., $k=\mp 2$