$\begin {array} {1 1} (A)\;\large\frac{130}{17\sqrt{29}} & \quad (B)\;\large\frac{13}{7\sqrt{29}} \\ (C)\;\large\frac{130}{7} & \quad (D)\;\text{None of these} \end {array}$

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- Distance of a point $(x_1,y_1)$ from the line $ax+by+c=0$ is $d=\bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$

Step 1 :

Let us find the point of intersection of the lines

$(3) 2x-3y+5=0$

$(2) 3x + 4y \qquad=0$

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$ \not6 \not x-9y+15=0$

$ \not 6 \not x +8y \qquad = 0$

$ (-) \quad (-)$

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$ \qquad -17y+15=0$

$ \therefore y=\large\frac{15}{17}$

and $x = \large\frac{20}{17}$

Hence the point of intersection is $ \bigg( \large\frac{20}{17}$$, \large\frac{15}{17} \bigg)$

Hence the distance of the line $5x-2y=0$ from the above point is

$ d = \bigg| \large\frac{5\bigg( \Large\frac{20}{17} \bigg)-2 \bigg( \Large\frac{15}{17} \bigg)}{\sqrt{5^2+2^2}} \bigg|$

$ = \bigg| \large\frac{\Large\frac{100}{17} - \Large\frac{30}{17}}{\sqrt{29}} \bigg|$

$ = \large\frac{130}{17\sqrt{29}}$

Hence option A is the correct answer.

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