The straight line which is both tangent and normal to the curve $x=3t^2,y=2t^3$ is:

Let line is tangent at $(3t_1^2,2t_1^3)$ and normal at   $(3t_2^2,2t_2^3)$
$\Rightarrow {\frac{dy}{dx}|}_{3t_1^2,2t_1^3}=t_1$ 
So, slope of normal at $(3t_2^2,2t_2^3)=-\frac{1}{t_2}$
$$\begin{gathered} \Rightarrow t_1=-\frac{1}{t_2} t_2=-\frac{1}{t_1} \\ \Rightarrow t_1=\frac{2t_1^3+\frac{2}{t_1^3}}{3(t_1^2-\frac{1}{t_1^2})} \\ \Rightarrow t_1^2(t_1^4-3)=2 or t_1\pm \sqrt{2} \end{gathered}$$