Question:

Step 1: The given equation $3x^2 - (1-3a)xy - ay^2 = 0$ represents a pair of straight lines passing through the origin $O(0,0)$. The points $A$ and $B$ are the intersections of the line $y = mx + 4$ with these two lines. Thus, $OA$ and $OB$ are the two lines represented by the given equation. Step 2: For $\angle AOB = 90^\circ$, the two lines represented by the equation $Ax^2 + Bxy + Cy^2 = 0$ must be perpendicular. The condition for perpendicularity is $A+C=0$. From the given equation, $A=3$ and $C=-a$. Therefore, $3 + (-a) = 0$, which implies $a = 3$. Step 3: Substitute $a=3$ into the equation of the curve: $$3x^2 - (1-3(3))xy - 3y^2 = 0$$ $$3x^2 - (1-9)xy - 3y^2 = 0$$ $$3x^2 + 8xy - 3y^2 = 0$$ Factor this quadratic equation to find the individual lines: $$(3x - y)(x + 3y) = 0$$ This gives two lines: $3x - y = 0 \Rightarrow y = 3x$ and $x + 3y = 0 \Rightarrow y = -\frac{1}{3}x$. The slopes of these lines are $k_1 = 3$ and $k_2 = -\frac{1}{3}$. Their product $k_1 k_2 = 3 \cdot (-\frac{1}{3}) = -1$, confirming that the lines are perpendicular. Step 4: The condition $\angle AOB = 90^\circ$ holds for all $m \in R - \{m_1, m_2\}$. This means the line $y=mx+4$ must not be parallel to either of the lines $OA$ or $OB$. If $y=mx+4$ were parallel to $y=3x$, then $m=3$. If it were parallel to $y=-\frac{1}{3}x$, then $m=-\frac{1}{3}$. Thus, $m_1 = -\frac{1}{3}$ and $m_2 = 3$ (since $m_1 < m_2$). The sum $m_1 + m_2$ is: $$m_1 + m_2 = -\frac{1}{3} + 3 = \frac{-1+9}{3} = \frac{8}{3}$$ Step 5: Calculate the area of $\triangle AOB$ when $m=2$. With $a=3$, the lines $OA$ and $OB$ are $y=3x$ and $y=-\frac{1}{3}x$. The line $y=mx+4$ becomes $y=2x+4$. To find point $A$, intersect $y=3x$ with $y=2x+4$: $3x = 2x+4 \Rightarrow x=4$. Substitute $x=4$ into $y=3x \Rightarrow y=3(4)=12$. So $A=(4,12)$. To find point $B$, intersect $y=-\frac{1}{3}x$ with $y=2x+4$: $-\frac{1}{3}x = 2x+4 \Rightarrow -x = 6x+12 \Rightarrow -7x=12 \Rightarrow x=-\frac{12}{7}$. Substitute $x=-\frac{12}{7}$ into $y=2x+4 \Rightarrow y=2(-\frac{12}{7})+4 = -\frac{24}{7}+\frac{28}{7} = \frac{4}{7}$. So $B=(-\frac{12}{7}, \frac{4}{7})$. The lengths $OA$ and $OB$ are: $$OA = \sqrt{4^2 + 12^2} = \sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10}$$ $$OB = \sqrt{\left(-\frac{12}{7}\right)^2 + \left(\frac{4}{7}\right)^2} = \sqrt{\frac{144}{49} + \frac{16}{49}} = \sqrt{\frac{160}{49}} = \frac{4\sqrt{10}}{7}$$ Since $\angle AOB = 90^\circ$, the area of $\triangle AOB$ is $\frac{1}{2} \cdot OA \cdot OB$: $$\text{Area} = \frac{1}{2} \cdot (4\sqrt{10}) \cdot \left(\frac{4\sqrt{10}}{7}\right) = \frac{1}{2} \cdot \frac{16 \cdot 10}{7} = \frac{1}{2} \cdot \frac{160}{7} = \frac{80}{7} \text{ sq. units}$$