I need help with b, but I also need to confirm something about a.
Lets talk about b, lets call the column vectors of the transformation matrix $w_1, w_2, w_3$. I can already see that $w_3 = [1, 2, 2]^T$ or simply the norm. But I am struggling with $w_1$ and $w_2$. I got a hint that they are supposed to be parallel but I just simply can't see it.
Lets go to a). I have done it with two methods. One worked but the other didn't even though it should? Maybe I have been using the wrong method but lets talk.
The method that didn't work: I rewrote the plane equation in parameter form and got the 2 linearly independent vectors $v_1 = [-4,1,1]$ and $v_2 = [4,-1,-1]$. Then I projected the first column vector in the standard basis for $R^3$ where $u_1 = [1,0,0]^T$. Then I added the projection of $u_1$ onto $v_1$ plus the projection of $u_1$ onto $v_2$. I did that with all $u_2$ and $u_3$. Then I created the transformation matrix by simply putting in these 3 vectors? Is this an incorrect method? It is what we use when doing Gram-Schmidt method tho?
Observe that the vectors $v_1,v_2$ are not linearly independent !
A simple way to find linearly independent vectors is to find basis of the plane. A point $(x,y,z)$ is on the plane if and only if $x + 2y+2z = 0$. That is, $$(x,y,z) = (-2y-2z, y,z) = y(-2,1,0) + z(-2,0,1)$$ This gives us a basis $(-2,1,0)$ and$(-2,0,1)$ of the plane.
Now use Gram-Schmidt to get an orthogonal basis of the plane. Then compute the projections of $(x,y,z)$ on each basis vector in the orthogonal basis of the plane. Sum up these projections to find $T(x,y,z)$. Then find the matrix.
For b). If $T$ has that matrix representation in a basis $v_1,v_2,v_3$ then it means that $T(v_1) = v_1$, $T(v_2) = v_2$ and $T(v_3) = 0$. Since $T$ is the projection onto the plane then the conditions on $v_1,v_2$ mean exactly that $v_1$ and $v_2$ are linearly independent vectors which span the plane. Finally take $v_3$ a normal vector to the plane as you did.