$\begin {array} {1 1} (A)\;18x-12y-11=0& \quad (B)\;18x-12y+11=0 \\ (C)\;18x+12y+11=0 & \quad (D)\;-18x+12y+11=0 \end {array}$

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- The perpendicular distance between of a line from a point $(x_1, y_1)$ is $ d= \bigg| \large\frac{Ax_1+By_1+c}{\sqrt{A^2+B^2}} \bigg|$

The equations of the given lines are :

$ \qquad 9x+6y-7=0$--------(1) and

$ \qquad 3x+2y+6=0$-------(2)

Let $p(a,b)$ be the point which is equidistant from lines (1) and (2)

The perpendicular distance of $p(a,b)$ from line (1) is given by

$ d_1 = \large\frac{|9a+6b-7|}{\sqrt{9^2+6^2}}$

$ = \large\frac{|9a+6b-7|}{\sqrt{117}}$

$ = \large\frac{|9a+6b-7|}{3\sqrt{13}}$

Similarly the perpendicular distance of $p(a,b)$ from line (2) is

$ d_2 = \large\frac{(3a+2b+6)}{\sqrt{(3)^2+(2)^2}}$

$ =\large\frac{(3a+2b+6)}{\sqrt{13}}$

Since $p(a,b)$ is equidistant from line (1) and (2) we get, $d_1=d_2$

$ \therefore \large\frac{|9a+6b-7|}{3\sqrt{13}}$$ = \large\frac{|3a+2b+6|}{\sqrt{13}}$

$ \Rightarrow |9a+6b-7| = 3|3a+2b+6|$

(i.e., ) $(9a+6b-7) = \pm 3(3a+2b+6) $ ( Considering the positive values)

(i.e.,) $(9a+6b-7) = 9a+6b+18$

This is not possible.

Hence $ 9a+6b-7 = -3(3a+2b+6)$

$\Rightarrow 9a+6b-7=-9a-6b-18$

$\Rightarrow 18a+12b+11=0$

Hence the required equation of the line is $ 18x+12y+11=0$

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