Basis for orthogonal complement possibly with linear combinations

In $$\mathbb{R}^4$$, consider the subspace $$W = Span(u_1, u_2,u_3)$$ with

$$u_1 = (-1,1,0,0)$$

$$u_2 = (-1,0,1,0)$$

$$u_3 = (-1,0,0,1)$$

$$(a)$$ Use the Gram-Schmidt Process to construct an orthonormal basis for $$W$$.

(b) Which of the vectors $$a=(1,1,0,-1)$$, $$b=(1,1,1,1)$$ and $$c=(0,0,-1,1)$$ is a basis for $$W^{\perp}$$?

(c) Express the vector $$x=(0,0,0,4)$$ as a sum of a vector $$x_1 \epsilon W$$ and $$x_2 \epsilon W^{\perp}$$.

I already did part $$(a)$$. The orthonormal basis is:

$$\left\{\frac{1}{\sqrt{2}}(-1,1,0,0),\frac{1}{\sqrt{6}}(-1,-1,2,0),\frac{1}{\sqrt{12}}(-1,-1,-1,3)\right\}$$

How do I do part $$(b)$$ and $$(c)$$ though?

For part $$(b)$$, my idea was to put $$u_1,u_2,u_3$$ into a matrix augmented with $$0$$ since $$W^{\perp}$$ means checking the null space. Alternatively, my second idea was to take the dot product of $$a,b,c$$ with $$u_1,u_2,u_3$$ to see which one results in zero. Is that the right idea?

For part $$(c)$$, am I just supposed to express $$x$$ as a linear combination of some vector in $$W$$ and $$W^{\perp}$$?

Yes, the vectors in $$W^\perp$$ are orthogonal to every vector of $$W$$. So the basis vector of $$W^\perp$$ must be orthogonal to each basis vector of $$W$$. Only $$\bf b$$ satisfies this condition.
For part $$(b)$$, note that the projection of $$\bf x$$ on $$W^\perp$$ is given by $$\displaystyle\mathbf{x_2}=\frac{\langle\mathbf x,\mathbf b\rangle}{\langle\mathbf b,\mathbf b\rangle}\mathbf b=(1,1,1,1)$$. So $$\mathbf x_1=\mathbf{x-x_2}=(-1,-1,-1,3)$$ lies in $$W$$. $$\mathbf x=(-1,-1,-1,3)+(1,1,1,1)$$
(b) It's the vector $$b$$: it's the anloy of the theree vectors which is orthogonal to every $$u_i$$. Besides, $$\dim W=3\implies\dim W^\perp=1$$.
(c) If$$v_1=\frac{1}{\sqrt{2}}(-1,1,0,0),\ v_2=\frac{1}{\sqrt{6}}(-1,-1,2,0),\text{ and }v_3=\frac{1}{\sqrt{12}}(-1,-1,-1,3)$$then$$\langle x,v_1\rangle v_1+\langle x,v_2\rangle v_2+\langle x,v_3\rangle v_3=(-1,-1,-1,3).$$So$$x=\overbrace{x-(-1,-1,-1,3)}^{\in W^\perp}+\overbrace{(-1,-1,-1,3)}^{\in W}.$$