$$O P=O Q$$ ($$\triangle P Q R$$ is isosceles triangle)

Let slope of line $$O P \rightarrow m_1$$

So, equation $$\rightarrow y=m_1 x$$

$$\begin{aligned} &\begin{aligned} & \tan 45^{\circ}=\left|\frac{m_1-m_2}{m_1 m_2}\right| \\ & 1=\left|\frac{m_1+\frac{3}{4}}{1-\frac{3}{4} m_1}\right| \\ & \Rightarrow 1-\frac{3}{4} m_1=m_1+\frac{3}{4} \\ & \frac{1}{4}=\frac{7}{4} m_1 \Rightarrow m_1=\frac{1}{7} \end{aligned}\\ &\text { Equation } O P \rightarrow y=\frac{1}{7} x \end{aligned}$$

Point of intersection of $$O P$$ & line $$3 x+4 y=12$$ is $$P\left(\frac{84}{25}, \frac{12}{25}\right)$$

$$\begin{aligned} & \Rightarrow \quad O P^2=a^2=\left(\frac{84}{25}\right)^2+\left(\frac{12}{25}\right)^2=\frac{288}{25} \\ & \therefore \quad I=O P^2+P Q^2+Q O^2 \\ &=a^2+a^2+2 a^2 \\ &=4 a^2 \\ &=4 \times \frac{288}{25} \\ & I=46.08 \\ & {[I] }=46 \end{aligned}$$