Answer:
Option C
Explanation:
We know that point of intersection of two lines
i.e, $ax_{1}+by_{1}+c_{1}=0$ and $ ax_{2}+by_{2}+c_{2}=0$ is
$\left(\frac{b_{1}c-b_{2}c}{a_{1}b_{2}-a_{2}b_{1}},\frac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}\right)$
$\therefore$ point of intersection of two lines i.e,
5x-6y-1=0 and 3x+2y+5=0 is
$\left(\frac{-6\times5-2\times-1}{5\times2-3\times-6},\frac{-1\times3-5\times5}{5\times2-3\times-6}\right)$
= $\left(\frac{-30+2}{10+18},\frac{-3-25}{10+18}\right)$
= $\left(\frac{-28}{28},\frac{-28}{28}\right)=\left(-1,-1\right)$
and slope of line 3x-5+11=0 is m= $\frac{3}{5}$
Equation of the line passing through (-1,-1) and perpendicular to 3x-5y+11=0 is
(y+1)= $-\frac{5}{3} (x+1)$
$\Rightarrow $ $3(y+1)=-\frac{5}{3} (x+1)$
$\Rightarrow $ $3y+3=-5x-5$
$\Rightarrow $ $5x+3y+3+3+5=0$
$\Rightarrow $ $5x+3y+8=0$
