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(D) The given equation of the pair of straight lines is $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$. We can factorize this as: $2x^2 + x(6 - y) - (3y^2 - y -

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$\frac{65}{\sqrt{26}}$

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(D) The given equation of the pair of straight lines is $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$. We can factorize this as: $2x^2 + x(6 - y) - (3y^2 - y - 4) = 0$. Using the quadratic formula for $x$: $x = \frac{-(6-y) \pm \sqrt{(6-y)^2 + 4(2)(3y^2 - y - 4)}}{4} = \frac{y-6 \pm \sqrt{y^2 - 12y + 36 + 24y^2 - 8y - 32}}{4} = \frac{y-6 \pm \sqrt{25y^2 - 20y + 4}}{4} = \frac{y-6 \pm (5y-2)}{4}$. This gives two lines: $L_1: x = \frac{6y-8}{4} \Rightarrow 2x - 3y + 4 = 0$. $L_2: x = \frac{-4y-4}{4} \Rightarrow x + y + 1 = 0$. The length of the perpendicular from $(-1, 5)$ to $2x - 3y + 4 = 0$ is $P_1 = \frac{|2(-1) - 3(5) + 4|}{\sqrt{2^2 + (-3)^2}} = \frac{|-2 - 15 + 4|}{\sqrt{13}} = \frac{13}{\sqrt{13}} = \sqrt{13}$. The length of the perpendicular from $(-1, 5)$ to $x + y + 1 = 0$ is $P_2 = \frac{|-1 + 5 + 1|}{\sqrt{1^2 + 1^2}} = \frac{5}{\sqrt{2}}$. The product of the lengths is $P_1 	imes P_2 = \sqrt{13} 	imes \frac{5}{\sqrt{2}} = \frac{5\sqrt{13}}{\sqrt{2}} = \frac{5\sqrt{26}}{2} = \frac{65}{\sqrt{26}}$.

The product of the lengths of the perpendiculars drawn from the point $(-1, 5)$ to the pair of lines $2x^2 - xy - 3y^2 + 6x + y + 4 = 0$ is

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