The Gram-Schmidt process is a very useful method to convert a set of linearly independent vectors into a set of orthogonal (or even orthonormal) vectors, in this case we want to find an orthogonal basis $\{v_i\}$ in terms of the basis $\{u_i\}$. It is an inductive process, so first let's define:
$$
v_1 := u_1 = (1,1,1,1).
$$
Then, by Gram-Schmidt:
$$
\begin{align}
v_2 :&= u_2 - \frac{\langle u_1 , u_2\rangle}{\langle u_1 , u_1\rangle}u_1\\
&= u_2 - \frac{2+4+1+5}{4}u_1 = (2,4,1,5) - 3(1,1,1,1)\\ &= (-1,1,-2,2).
\end{align}
$$
and finally
$$
\begin{align}
v_3 &= u_3 - \frac{\langle v_2 , u_3\rangle}{\langle v_2 , v_2\rangle}v_2 - \frac{\langle v_1 , u_3\rangle}{\langle v_1 , v_1\rangle}v_1\\
&= u_3 + \frac{30}{10}v_2 + \frac{8}{4}v_1 \\
&= (1,-5,4,-8) + 3(-1,1,-2,2) + 2(1,1,1,1) \\
&= (0,0,0,0)
\end{align}
$$
Now,why are we getting zero? what does this mean!? Basically, getting zero in the last result means your original vectors aren't linearly independent (we should've checked this before doing Gram-Schmidt, but I didn't think about it)! Some basic calculations show that $u_3 -7u_1 + 3u_2 =0$.
Still the previous results work, since $v_1$ and $v_2$ are orthogonal and span $W$ (why?).