= Solution
{c}
Let
$$
\mathbf r_0=\frac{\mathbf a\wedge\mathbf b}{|\mathbf a|^2},\qquad
\mathbf N=\mathbf b+\mathbf a\wedge\mathbf y.
$$
The <vector triple product> identity $(\mathbf a\wedge\mathbf b)\wedge\mathbf a=|\mathbf a|^2\mathbf b-(\mathbf a\cdot\mathbf b)\mathbf a$ shows that $\mathbf r_0\wedge\mathbf a=\mathbf b$. Moreover, $(\mathbf r-\mathbf r_0)\wedge\mathbf a=\mathbf0$ exactly when $\mathbf r-\mathbf r_0$ is a <scalar multiple> of $\mathbf a$. Thus the given <line> is
$$
L=\{\mathbf r_0+t\mathbf a:t\in\mathbb R\}.
$$
Also $\mathbf a\wedge\mathbf r_0=-\mathbf b$, so $\mathbf N=\mathbf a\wedge(\mathbf y-\mathbf r_0)$. The assumption $\mathbf N\ne\mathbf0$ says precisely that $\mathbf y$ is outside $L$. Consequently $L$ and $\mathbf y$ determine a unique <plane>, with <normal vector> $\mathbf N$.
To identify its equation, use cyclic invariance of the <scalar triple product>:
$$
\mathbf r\cdot(\mathbf a\wedge\mathbf y)
=\mathbf y\cdot(\mathbf r\wedge\mathbf a).
$$
For $\mathbf r\in L$, this equals $\mathbf y\cdot\mathbf b$, while $\mathbf r\cdot\mathbf b=\mathbf r\cdot(\mathbf r\wedge\mathbf a)=0$. Hence $\mathbf r\cdot\mathbf N=\mathbf y\cdot\mathbf b$ throughout $L$. The <point> $\mathbf y$ also satisfies that equation because $\mathbf y\cdot(\mathbf a\wedge\mathbf y)=0$. Since $\mathbf N\ne\mathbf0$, the required <plane> is therefore exactly
$$
\boxed{\mathbf r\cdot(\mathbf b+\mathbf a\wedge\mathbf y)=\mathbf y\cdot\mathbf b.}
$$
This also proves the reusable <plane through a point and a line given by a cross product> construction.
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