Let $$U= \operatorname{span}\{(1,0,0)\} \\ V = \operatorname{span}\{(0,1,1)\} \\ W=\operatorname{span}\{(0,1,0),(0,0,1)\}$$
Then clearly every vector in $U$ is orthogonal to every vector in $V$ and to every vector in $W$. Hence $U$ and $V$ are orthogonal subspaces and $U$ and $W$ are orthogonal subspaces.
But $U$ and $V$ together don't make up all of the vectors in $\Bbb R^3$ (the ambient space). For instance $(0,0,1)$ is not in either one. However $U$ and $W$ together do in fact make up all of the vectors in $\Bbb R^3$. In the lingo of linear algebra, we say that the direct sum of $U$ and $W$ is $\Bbb R^3$ -- or symbolically $U\oplus W = \Bbb R^3$.$^\dagger$ It's this extra property that makes $U$ and $W$ not just orthogonal subspaces -- but each other's orthogonal complement.
$\dagger$: I'm being a little loose with my language here. Technically the condition for $U\oplus W=\Bbb R^3$ (which is the required extra condition) is that every vector in $\Bbb R^3$ can be written in a unique way as a sum of an element of $U$ and of $W$. I.e. for every $v\in \Bbb R^3$, there exists a unique $u\in U$ and a unique $w\in W$ such that $u+w=v$. Nevertheless, the statements I give in the paragraph above are how we intuitively think of orthogonal complements.
An alternative approach is, as you mention, to consider the vector $b$ and an orthogonal vector to $b$ in the same plane as $a, b$ and $c$. You can build one by considering the part of $a$ orthogonal to $b$, say
\begin{equation}
d = a - \frac{a \cdot b}{b \cdot b}b \, .
\end{equation}
Now you can expand $c$ as a linear combination of $b$ and $d$, say
\begin{equation}
c = \beta \, b + \delta \, d \, .
\end{equation}
Then the one additional constraint, namely the fact that $c$ is at an angle $\phi$ with $b$, can be written as
\begin{equation}
\frac{b \cdot c}{\sqrt{b^2 \, c^2}} = \cos \phi \,.
\end{equation}
Out of the two unknowns, $\beta$ and $\delta$, we will be able to fix one of them in terms of the other (up to a sign) with this constraint. The sign arbitrariness is simply due to the fact that there are generically two distinct co-planar rays at an angle $\phi$ with $b$: one on the "right", one on the "left". The clear exceptions are when $\phi$ is an integer multiple of $\pi$, in which case there is only a single distinct ray. We can write the solution in a way that covers these cases as well,
\begin{equation}
\delta \, |\cos \phi| = \pm \, \lambda \, \beta \,|\sin \phi| \, ,
\end{equation}
with
\begin{equation}
\lambda = \frac{b^2} {\sqrt{a^2 b^2 - \left( a \cdot b \right)^2}} \, .
\end{equation}
EDIT: Let's see how to find this relation. First, remember that $d$ and $b$ are orthogonal. Hence, $b \cdot d = 0$ and then $b \cdot c = \beta \, b^2$. Orthogonality also implies that $c^2 = \beta^2 \, b^2 + \delta^2 \, d^2$. By taking this into account and squaring the constraint for $\cos \phi$, you'll find $\beta^2 \, b^2 \, \sin^2 \phi = \delta^2 \, d^2 \, \cos^2 \phi$. Then if you expand $d^2$, you'll find the relation above.
Best Answer
I don't like this line for several reasons, the two main reasons being that
This is absolutely correct. The orthogonal space is a collection of vectors, not a collection of planes. And all vectors are rooted at the origin. So the orthogonal space consists of all vectors in the plane containing the origin which has normal vector $v$ (this is essentially the definition of the normal vector of a plane).