(a) To find a basis for the plane $x-y+z = 0$, you could solve this equation in terms of $x$ and $z$: $y= x+z$. Then the set of vectors of your plane could be described as:
$$
V = \left\{ (x, x+z, z) \ \vert \ x,z \in \mathbb{R} \right\} \ .
$$
From this description it's easy to find a basis for your plane $V$: it will have two vectors, say $u,v \in \mathbb{R}^3$:
$$
V = [u,v] \ .
$$
Then, the orthonormal subspace to your plane is a straight line, generated by the cross product of $u$ and $v$
$$
V^\bot = [u \times v] \ .
$$
(b) The closest point $\hat{b}$ to $b$ in the plane $V$ is the orthogonal projection of $b$ onto $V$. In this case, you don't really need the formulae you'll see in Wikipedia, because it's just the intersection of $V$ with the straight line with direction $u\times v$ that passes through $b$. That is, you have to solve the system of linear equations
$$
\begin{align}
x - y +z &= 0 \\
(x,y,z) &= (1,2,0) + \lambda u\times v \ .
\end{align}
$$
(c) The "error" between $b$ and $\hat{b}$ is the same as the distance from $b$ to $\hat{b}$; that is, $\| b - \hat{b} \|$.