Projection of the vector $\vec{b}=\left(\begin{smallmatrix}3\\-1\\4\end{smallmatrix}\right)$ in some given direction 
Find the projection of the vector $\vec{b}=\begin{pmatrix}
3 \\ -1\\ 4
\end{pmatrix}$ in the direction
$(a)$ perpendicular to the plane: $$2x-2y+z=5.$$
  $(b)$ parallel to the plane: $$2x-2y+z=5.$$

For the part $(a)$ projecting $\vec{b}=\begin{pmatrix}3 \\ -1\\ 4\end{pmatrix}$ to $\vec{a}=\begin{pmatrix}2 \\ -2\\ 1\end{pmatrix}$ will give me the perpendicular projection right?
For part $(b)$ I know I have to subtract what I found in a from a vector to find the projection of the vector parallel to the plane, but I'm kinda lost.

Perpendicular 
The plane $E = 2x-2y+z=5 $ is defined by the normal vector $\vec{n}=\begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix}$ and the projection of $\vec{b}$ on $\vec{n}$, we define it as $\vec{l}$,  is obtained by $\,|l|\cdot \left(\frac{1}{|n|}\vec{n}\right)\,$   (See unit vector)
It follows:
$$|l| = |b|\cdot \cos(\alpha) = |b|\frac{\vec{b}\cdot \vec{n}}{|b|\cdot |n|} = 
\frac{\vec{b}\cdot \vec{n}}{|n|} = 4,\quad\text{ and }\quad \frac{1}{|n|}\vec{n} = \frac{1}{3}\cdot\begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix}$$
$$\vec{l} = |l|\cdot \left(\frac{1}{|n|}\cdot\vec{n}\right) = 4\cdot\frac{1}{3}\cdot \begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix} = \begin{pmatrix}{\frac{8}{3} \\ -\frac{8}{3}\\ \frac{4}{3}}\end{pmatrix}$$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$

Parallel
And the vector $\vec{t}$ ist parallel to the plane $E$. This is also the projection of $\vec{b}$ on $E$:
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
$$\vec{t} = \vec{b}-\vec{l}  = \begin{pmatrix}{\quad 3 - \frac{8}{3}\\ -1 + \frac{8}{3}\\\quad 4 - \frac{4}{3}}\end{pmatrix} = \begin{pmatrix}{\frac13\\\frac53\\\frac83 }\end{pmatrix}
$$
