First, row reduce A, find general solution to Ax = 0, and thus find basis for N(A). I found parametric solution is [-5t,7t-s,s,t] and thus basis is [-5,7,0,1] and [0,-1,1,0].
Put basis as columns in a matrix B.
The orthogonal projection matrix is P = B(B^T B)^-1 B^T.
The orthogonal projection is v = Py = [5,-5,-2,-1]. Check this is in N(A) since Av = 0 and/or v is linear combination of basis vectors.
Check y - v = [2,1,1,3] is in A, so is orthogonal to N(A) and thus v is indeed an orthogonal projection of y.