$\begin{aligned}
& \Rightarrow \text { In } \triangle \mathrm{ABC} \text {, } \\
& \frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9} \Rightarrow \mathrm{k} \\
& \therefore \quad \mathrm{a}+\mathrm{b}=7 \mathrm{k} ...(i)\\
& \mathrm{b}+\mathrm{c}=8 \mathrm{k} ... (ii) \\
& \mathrm{c}+\mathrm{a}=9 \mathrm{k} ... (iii)\\
&
\end{aligned}$
Adding above equations,
$\begin{aligned}
& 2 a+2 b+2 c=24 k \\
& a+b+c=12 k ... (iv)
\end{aligned}$
Solving equations (i), (ii), (iii), (iv)
We get,
$\begin{array}{ll}
& \mathrm{c}=5 \mathrm{k}, \mathrm{a}=4 \mathrm{k}, \mathrm{b}=3 \mathrm{k} \\
\therefore \quad & \mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2
\end{array}$
$\therefore \quad \triangle \mathrm{ABC}$ is right angled triangle
$\therefore \quad \angle \mathrm{C}=90^{\circ}$


$\begin{aligned} & \text { Area of } \begin{aligned} \triangle \mathrm{ABC} & =\frac{1}{2} \mathrm{ab} \sin \mathrm{C} \\ & =\frac{1}{2} \mathrm{ab} \sin 90 \\ & =\frac{1}{2} \times 4 \mathrm{k} \times 3 \mathrm{k} \\ & =6 \mathrm{k}^2\end{aligned} \\ & \therefore \quad \text { Now, } \frac{[\mathrm{A}(\Delta \mathrm{ABC})]^2}{\mathrm{k}^4}=\frac{\left(6 \mathrm{k}^2\right)^2}{\mathrm{k}^4}=\frac{36 \mathrm{k}^4}{\mathrm{k}^4}=36\end{aligned}$