Given pair of lines is $8 x^2-6 x y+y^2=0$.
$\begin{aligned} & \Rightarrow \quad 8\left(\frac{x}{y}\right)^2-6\left(\frac{x}{y}\right)+1=0 \\ & \Rightarrow \quad\left[4\left(\frac{x}{y}\right)-1\right]\left[2\left(\frac{x}{y}\right)-1\right]=0 \\ & \Rightarrow \quad \frac{4 x}{y}=1 \text { and } \frac{2 x}{y}=1 \\ & \Rightarrow \quad 4 x=y \quad \text { and } \quad 2 x=y \\ & \end{aligned}$
Also, other line is $2 x+3 y=a$
On solving Eqs. (i) and (ii), we get
$O(0,0), A\left(\frac{a}{14}, \frac{2 a}{7}\right)$ and $B\left(\frac{a}{8}, \frac{a}{4}\right)$
$\begin{array}{ll}\therefore \text { Area of } \triangle A B C=\frac{1}{2}|| \begin{array}{ccc}0 & 0 & 1 \\ \frac{a}{14} & \frac{2 a}{7} & 1 \\ \frac{a}{8} & \frac{a}{4} & 1\end{array} \| \\ \Rightarrow & 7=\frac{1}{2}\left|\left(\frac{a^2}{56}-\frac{2 a^2}{56}\right)\right| \\ \Rightarrow & 7=\frac{1}{2}\left|\frac{-a^2}{56}\right| \\ \Rightarrow & 14 \times 56=a^2 \\ \Rightarrow & a=28\end{array}$