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(C) The given equation is a homogeneous equation of the third degree,which represents three straight lines passing through the origin.Let the three li

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$a^2 + ac + bd + d^2 = 0$

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(C) The given equation is a homogeneous equation of the third degree,which represents three straight lines passing through the origin.Let the three lines be $L_1, L_2, L_3$. Suppose $L_1$ and $L_2$ are perpendicular.The equation of a pair of perpendicular lines passing through the origin is of the form $x^2 + pxy - y^2 = 0$ (assuming the lines are $y = m_1x$ and $y = m_2x$ with $m_1m_2 = -1$).Thus,we can write the cubic equation as: $(x^2 + pxy - y^2)(ax - dy) = 0$.Expanding this,we get: $ax^3 - dx^2y + apxy^2 - dp xy^2 - axy^2 + dy^3 = 0$,which simplifies to $ax^3 + (ap - d)x^2y + (ap - d)xy^2 + dy^3 = 0$ is incorrect; let us re-expand: $(x^2 + pxy - y^2)(ax - dy) = ax^3 - dx^2y + apxy^2 - dp xy^2 - axy^2 + dy^3 = ax^3 + (ap - d)x^2y + (-a - dp)xy^2 + dy^3 = 0$.Comparing coefficients with $ax^3 + bx^2y + cxy^2 + dy^3 = 0$:$b = ap - d \Rightarrow ap = b + d$$c = -a - dp \Rightarrow dp = -c - a$From the first,$p = \frac{b+d}{a}$. Substituting into the second: $d(\frac{b+d}{a}) = -c - a$.$bd + d^2 = -ac - a^2$.Rearranging gives $a^2 + ac + bd + d^2 = 0$.

If two of the three lines represented by the equation $ax^3 + bx^2y + cxy^2 + dy^3 = 0$ are perpendicular,then:

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