Find k, if the line 2x - 3y = 11 is perpendicular to the line 3x + ky = -4?

Answer: Option D

$$\begin{array}{rl}& 2x-3y=11\\ & 3y=2x-11\\ & y=\frac{2}{3}x-\frac{11}{3}\\ & \left(y=mx+c,\text{ where }m\text{ is slope}\right)\\ & \text{Slope}=m_1=\frac{2}{3}\\ & 3x+ky=-4\\ & ky=-3x-4\\ & y=-\frac{3}{k}x-\frac{4}{k}\\ & m_2=-\frac{3}{k}\end{array}$$
Relation between slope of perpendicular lines
$$\begin{array}{rl}& m_1{m}_2=-1\\ & \Rightarrow \left(\frac{2}{3}\right)\times \left(-\frac{3}{k}\right)=1\\ & \Rightarrow k=2\end{array}$$