= Solution
{c}
The three <lines> have the convenient <parametric equations>
$$
L_1:\ (s,0,1),\qquad L_2:\ (1,s,0),\qquad L_3:\ (0,1,s),\qquad s\in\mathbb R.
$$
For $L_1$, substitute $\mathbf a=\mathbf i$, $\mathbf b=\mathbf j$ in the <plane through a point and a line given by a cross product> formula. Its <normal vector> is $\mathbf j+\mathbf i\wedge\mathbf y=(0,1-y_3,y_2)$. Cyclically permuting the <coordinates> gives all three <planes>:
$$
\boxed{\begin{aligned}
\Pi_1:&\quad(1-y_3)r_2+y_2r_3=y_2,\\
\Pi_2:&\quad y_3r_1+(1-y_1)r_3=y_3,\\
\Pi_3:&\quad(1-y_2)r_1+y_1r_2=y_1.
\end{aligned}}
$$
Each <normal vector> is nonzero because $\mathbf y$ is not on the corresponding <line>. Thus these are genuine <planes>, and each is unique.
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