Well, V3 is amultipleof that vector- you know the direction but not the length. But any multiple will work here, so, yes, (13, 9, -4) is good.

This is aprojectionoperator that maps any 3 vector onto the subspace spanned by V1 and V2.

If you were to use V1, V2, and V3 as basis vectors you would get the matrix

.

If you mean for the matrix in the "standard" basis, (1, 0, 0), (0, 1, 0), and (0, 0, 1), note that the matrix

having V1, V2, and V3 as columns is the "change of basis matrix"- it maps the standard basis into the new basis. Of course, then, goes the other way. That means the matrix product maps a vector in the standard basis to the new basis, then applies the operator A, then goes back to the standard basis- it gives the matrix for A in the standard basis.