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If $(h, k)$ is the image of the point $(3,-4)$ with respect to the line $2 x-3 y-5=0$ and $(\ell, \mathrm{m})$ is the foot of the perpendicular from $(h, k)$ on to the line $3 x+2 y+$ $12=0$, then $\ell \mathrm{h}+\mathrm{mk}+1=$
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$5$
$\frac{h-3}{2}=\frac{k+4}{-3}=\frac{-2(3.2-3(-4)-5)}{2^2+(-3)^2}$
$\begin{aligned} & \frac{h-3}{2}=\frac{k+4}{-3}=\frac{-2 \times 13}{13} \therefore \frac{h-3}{2}=-2 \text { and } \frac{k+4}{-3}=-2 \\ & \mathrm{~h}=-1 \text { and } \mathrm{k}=2 \\ & \frac{l+1}{3}=\frac{m-2}{2}=\frac{-(3(-1)+2(2)+12)}{3^2+2^2} \\ & \frac{l+1}{3}=\frac{m-2}{2}=\frac{-13}{13}=-1\end{aligned}$
$\begin{aligned} & l=-4, m=0 \\ & \therefore l h+m k+1=-4(-1)+(0)(2)+1=5\end{aligned}$
$\begin{aligned} & \frac{h-3}{2}=\frac{k+4}{-3}=\frac{-2 \times 13}{13} \therefore \frac{h-3}{2}=-2 \text { and } \frac{k+4}{-3}=-2 \\ & \mathrm{~h}=-1 \text { and } \mathrm{k}=2 \\ & \frac{l+1}{3}=\frac{m-2}{2}=\frac{-(3(-1)+2(2)+12)}{3^2+2^2} \\ & \frac{l+1}{3}=\frac{m-2}{2}=\frac{-13}{13}=-1\end{aligned}$
$\begin{aligned} & l=-4, m=0 \\ & \therefore l h+m k+1=-4(-1)+(0)(2)+1=5\end{aligned}$
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