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(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$. We can rewrite the quadratic part as $2(4x^2 - 12xy + 9y^2) = 2(2x - 3y)^2$. Let th

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$\frac{7}{2\sqrt{13}}$

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(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$. We can rewrite the quadratic part as $2(4x^2 - 12xy + 9y^2) = 2(2x - 3y)^2$. Let the lines be of the form $(2x - 3y + c_1)(2x - 3y + c_2) = 0$. Expanding this,we get $4x^2 - 12xy + 9y^2 + 2(c_1 + c_2)x - 3(c_1 + c_2)y + c_1c_2 = 0$. Comparing this with the given equation $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$,we divide the given equation by $2$: $4x^2 - 12xy + 9y^2 - 3x + 4.5y - 2.5 = 0$. Here,$c_1 + c_2 = -3/2$ and $c_1c_2 = -2.5$. The distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Using $(c_1 - c_2)^2 = (c_1 + c_2)^2 - 4c_1c_2 = (-1.5)^2 - 4(-2.5) = 2.25 + 10 = 12.25$. So,$|c_1 - c_2| = \sqrt{12.25} = 3.5 = 7/2$. The distance is $\frac{7/2}{\sqrt{2^2 + (-3)^2}} = \frac{7/2}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.

The distance between the two parallel lines represented by the equation $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$ is

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