Find the point on the curve where the plane perpendicular to the curve is parallel a plane 
Find the point on the curve $\langle t^3, 3t, t^4 \rangle$ where the plane perpendicular to the curve is parallel to the plane $24x + 6y – 64z = 0$

I assume to find this point I need to parameterize this equation with the curve?
$x \Rightarrow t^3=24u \Rightarrow u = \frac{24}{t^3}$
$y \Rightarrow 3t=6u \Rightarrow u = \frac{t}{2}$
$z \Rightarrow t^4=-64u \Rightarrow u = \frac{-t^4}{64}$
Am I on the right track here?
 A: Hint:
The tangent vector at $\bigl(x(t_0),y(t_0),z(t_0)\bigr)$ is normal to the plane $24x-6y+64z=0$.
So you have to solve for $\;\dfrac{3t_0^2}{24}=\dfrac{3}{6}=\dfrac{4t_0^3}{-64}$.
A: Try to interpret each of the following steps for yourself.
$$\left\{ {\begin{array}{*{20}{l}}
{x = {t^3}}\\
{y = 3t}\\
{z = {t^4}}
\end{array}} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop  \to \limits^{{\rm{1st}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}
{\frac{{dx}}{{dt}} = 3{t^2}}\\
{\frac{{dy}}{{dt}} = 3}\\
{\frac{{dz}}{{dt}} = 4{t^3}}
\end{array}} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop  \to \limits^{{\rm{2nd}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}
{3{t^2} = \alpha .24}\\
{3 = \alpha .6}\\
{2{t^3} = \alpha . - 64}
\end{array}} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop  \to \limits^{{\rm{3rd}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}
{\alpha  = \frac{1}{2}}\\
{t =  - 2}
\end{array}} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop  \to \limits^{{\rm{4th}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left\{ {\begin{array}{*{20}{l}}
{x =  - 8}\\
{y =  - 6}\\
{z = 16}
\end{array}} \right.$$
