Let \(a, b, c\) be the lengths of sides of triangle \(\mathrm{ABC}\) such that \(\frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9}=\mathrm{k}\). Then \(\frac{(\mathrm{A}(\triangle \mathrm{ABC}))^2}{\mathrm{k}^4}=\)

\(\begin{aligned}
& \Rightarrow \text { In } \triangle \mathrm{ABC} \text {, } \\
& \frac{\mathrm{a}+\mathrm{b}}{7}=\frac{\mathrm{b}+\mathrm{c}}{8}=\frac{\mathrm{c}+\mathrm{a}}{9} \Rightarrow \mathrm{k} \\
& \therefore \quad \mathrm{a}+\mathrm{b}=7 \mathrm{k} ...(i)\\
& \mathrm{b}+\mathrm{c}=8 \mathrm{k} ... (ii) \\
& \mathrm{c}+\mathrm{a}=9 \mathrm{k} ... (iii)\end{aligned}\)
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,
\(\mathrm{c}=5 \mathrm{k}, \mathrm{a}=4 \mathrm{k}, \mathrm{b}=3 \mathrm{k}\)
\(\therefore \mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2\)
\(\therefore \triangle \mathrm{ABC}\) is right angled triangle
\(\therefore \angle \mathrm{C}=90^{\circ}\)

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