If $3x–ky–2=0,2x+y+2=0$  are perpendicular, then the value of $‘k’$  is

$a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$

$m_1=\frac{-3}{-k},m_2=\frac{-2}{1}$

Given lines are perpendicular

$m_1\times m_2=-1$

$\frac{3}{k}\times -2=-1$

$6=k$

$\therefore$ $k=6$