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(C) The given equation is $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$.First,we factorize the homogeneous part $6x^2 - 5xy - 6y^2$:$6x^2 - 9xy + 4xy - 6y^2 =

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$\frac{1}{13}$

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(C) The given equation is $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$.First,we factorize the homogeneous part $6x^2 - 5xy - 6y^2$:$6x^2 - 9xy + 4xy - 6y^2 = 3x(2x - 3y) + 2y(2x - 3y) = (3x + 2y)(2x - 3y)$.Let the lines be $(3x + 2y + c_1)(2x - 3y + c_2) = 0$.Comparing the coefficients with $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$:$c_2(3x + 2y) + c_1(2x - 3y) = x + 5y \implies (3c_2 + 2c_1)x + (2c_2 - 3c_1)y = x + 5y$.Solving $3c_2 + 2c_1 = 1$ and $2c_2 - 3c_1 = 5$,we get $c_1 = -1$ and $c_2 = 1$.Thus,the lines are $3x + 2y - 1 = 0$ and $2x - 3y + 1 = 0$.The perpendicular distance from $(0, 0)$ to $3x + 2y - 1 = 0$ is $d_1 = \frac{|-1|}{\sqrt{3^2 + 2^2}} = \frac{1}{\sqrt{13}}$.The perpendicular distance from $(0, 0)$ to $2x - 3y + 1 = 0$ is $d_2 = \frac{|1|}{\sqrt{2^2 + (-3)^2}} = \frac{1}{\sqrt{13}}$.The product of the distances is $d_1 	imes d_2 = \frac{1}{\sqrt{13}} 	imes \frac{1}{\sqrt{13}} = \frac{1}{13}$.

The product of the perpendicular distances drawn from the origin to the pair of straight lines $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$ is

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