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If the equation of a line parallel to $3 x-2 y+5=0$ and at a distance of 5 units from it is $3 x-2 y+C=0$, then $C$ is equal to
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$5( \pm \sqrt{13}+1)$
$\begin{aligned} &\frac{|c-5|}{\sqrt{9+4}}=5 \Rightarrow|c-5|=5 \sqrt{13} \\ & \Rightarrow \quad c-5= \pm 5 \sqrt{13} \\ & \Rightarrow \quad c=5 \pm 5 \sqrt{13}=5(1 \pm \sqrt{13})\end{aligned}$
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