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(B) The given equation of the pair of straight lines is $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$. For a general equation of a pair of lines $Ax^2 + 2Hxy + By^

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$2$

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(B) The given equation of the pair of straight lines is $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$. For a general equation of a pair of lines $Ax^2 + 2Hxy + By^2 = 0$,the lines are perpendicular if the sum of the coefficients of $x^2$ and $y^2$ is zero,i.e.,$A + B = 0$. Here,$A = 3d$ and $B = d^2 - 2$. Setting $A + B = 0$,we get $3d + d^2 - 2 = 0$,which is $d^2 + 3d - 2 = 0$. Using the quadratic formula $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we find $d = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2}$. Since there are two distinct real values for $d$,the condition is satisfied for $2$ values of $d$. Thus,option $B$ is correct.

The pair of straight lines is represented by the equation $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$. If the lines are perpendicular to each other,for how many values of $d$ will this condition be satisfied?

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