$A=\begin{bmatrix} 1 & -3 & -3\\ 1 & 5 &1 \\ 1& 7 & 2 \end{bmatrix}$ , $\vec{b}=\begin{bmatrix} 5\\ -3\\ -5 \end{bmatrix}$
Part (c) of this question is what I am struggling with. Here is the rest of the question for context.
(a) Find an orthogonal basis for Col(A).
I have used the Gram-Schmidt process to create the set of vectors:
$\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ , $\begin{bmatrix} -6\\ 2\\ 4 \end{bmatrix}$ , $\begin{bmatrix} -18\\ 6\\ 12 \end{bmatrix}$.
(b) Find a basis for Nul(A)
I set up $A\vec{x}=0$, and got the following as the basis: $\begin{bmatrix} 3\\ 1\\ 0 \end{bmatrix}$, $\begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix}$.
(c) Find the projection $\hat{b}$ of $\vec{b}$ onto Col(A).
Now, this is where I'm getting thrown off. Not because I do not understand how to do this calculation, but because I'm not sure what I am projecting onto. What I know is that Col(A) is the set of all linear combinations of the columns of $A$. So, at first glance I thought to project $\vec{b}$ onto the columns of $A$. However, part of me thinks I should project $\vec{b}$ onto the orthogonal basis for Col(A), because that was the first part of this question.
Am I overthinking what part (c) is asking?