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(D) Given line $2 x+3 y=k \Rightarrow \frac{2 x+3 y}{k}=1 \dots (i)$Also given,$3 x^2-x y+3 y^2+2 x-3 y-4=0$Now homogenizing the above equation using

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$6 k^2-5 k-52=0$

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(D) Given line $2 x+3 y=k \Rightarrow \frac{2 x+3 y}{k}=1 \dots (i)$Also given,$3 x^2-x y+3 y^2+2 x-3 y-4=0$Now homogenizing the above equation using $(i)$:$3 x^2-x y+3 y^2+(2 x-3 y)\left(\frac{2 x+3 y}{k}\right)-4\left(\frac{2 x+3 y}{k}\right)^2=0$Multiplying by $k^2$:$k^2(3 x^2-x y+3 y^2) + k(4 x^2+6 x y-6 x y-9 y^2) - 4(4 x^2+9 y^2+12 x y) = 0$$x^2(3 k^2+4 k-16) + x y(-k^2-48) + y^2(3 k^2-9 k-36) = 0$Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(3 k^2+4 k-16) + (3 k^2-9 k-36) = 0$$6 k^2-5 k-52 = 0$

If the lines joining the origin to the points of intersection of $2 x+3 y=k$ and $3 x^2-x y+3 y^2+2 x-3 y-4=0$ are at right angles,then

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