Finding Projection Vector Find the projection of the vector $t = [3, 3, 3]^T$ onto the subspace spanned by the vectors $\{x, y\}$, where $x = [6; 1; -3]$, $y = [1; 0; 2]$.
I was told to look at the orthogonal basis, project the vector to each basis element, and then add them up. So do I use the projection formula for $t$ and $x$, and $t$ and $y$, and then add those two together?

The method you state is good and is basically what the other two answers recommend.
However, this involves calculating two projections. This is not onerous, but there is an easier way to project a 3-D vector onto a 2-D plane defined by two linearly independent vectors.
1) Find the cross-product of the two vectors that define the plane.
2) Find the projection of your vector onto that vector.
3) Subtract that projection from your original vector.
You now have the projection of the original vector onto the plane. This took only one projection calculation, one cross-product, and one subtraction. The other method takes two projection calculations and an addition. Since cross-products are easier than projections, this method is easier. Note also that this method did not even require that the two basis vectors of the plane are orthogonal to each other--just that they are linearly independent.
