Let the equation of line forming a triangle having area $12 \mathrm{sq}$. units is
$$
\frac{x}{a}+\frac{y}{b}=1
$$

So,
$$
|a b|=24
$$
$\because$ Line (i) passes through point $(2,3)$, so
$$
\frac{2}{a}+\frac{3}{b}=1
$$
from Eqs. (ii) and (iii), we get
$$
|2 b+3 a|=24
$$

If $a$ and $b$ are positive, then
$$
\begin{array}{rlrl}
& & a b=24 \text { and } 3 a+2 b=24 \\
\Rightarrow & a\left(\frac{24-3 a}{2}\right)=24 \\
\Rightarrow & 3 a^2-24 a+48=0 \\
& \because \text { Discriminant }=24^2-4 \times 3 \times 48=0 \\
\therefore & a=4 \text { and } b=6
\end{array}
$$

If $a$ is positive and $b$ is negative then $a b=-24$ and $3 a+2 b=24$.
$$
\Rightarrow \quad 2\left(\frac{24-3 a}{2}\right)=-24 \Rightarrow 3 a^2-24 a-48=0
$$
$\because$ Discriminant $=24^2+4 \times 3 \times 48>0$


$$
\begin{array}{ll}
\therefore \quad a=\frac{24 \pm \sqrt{2 \times 24^2}}{6}=4+4 \sqrt{2} \quad \because a>0 \\
\text { and } \quad b=6-6 \sqrt{2}
\end{array}
$$

If $a$ is negative and $b$ is positive, then $a b=-24$ and $3 a+2 b=24$
$$
\Rightarrow \quad a=4-4 \sqrt{2} \text { and } b=6+6 \sqrt{2}
$$

And at the last it is not possible that $a$ and $b$ both are negative.
So, 3 triangles are possible.