I can't see how to do this. I cant see how those two vectors can span a subspace of $\mathbb{R}^3$ when they both have 4 components.
Here's what I've done. I tried to find the column space of these two vectors by putting them into a $4\times2$ matrix
$ \left[\begin{array}{rr|r} -4 & 2 & a \\ 2 & 2 & b \\ 2 & 2 & c \\ -4 & 0 & d \end{array}\right] $
I then row-reduced and got the following results.
$ \left[\begin{array}{rr|l} -4 & 2 & a \\ 0 & 6 & 2b - a\\ 0 & 0 & c - b \\ 0 & 0 & 3d - 2a + 2b \end{array}\right] $
So then I used the equation
$3d - 2a + 2b = 0$ to obtain two $1\times3$ vectors that span $\mathbb{R}^3$
$a = b + \frac{3}{2}d$
$\begin{bmatrix} a \\ b \\ d \end{bmatrix} = \operatorname{span} \left\{\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} \frac{3}{2} \\ 0 \\ 1 \end{bmatrix}\right\}$
But I cant project a $4\times1$ onto these vectors because I cant dot a $1\times4$ vector with a $1\times3$ vector.
I have a feeling I have got the workings of this question all wrong. What am I supposed to be doing?