= Solution
In the parametric description, $a$ is a point of the <straight line> and the nonzero <vector> $b$ gives its direction. The real parameter runs through every point of that line. For the <cross product> description, take another <cross product> with $c$ and use the <vector triple product identity>:
$$
c\times(x\times c)=|c|^2x-(c\cdot x)c=c\times d.
$$
Hence every solution has the representation
$$
\boxed{x=a+\lambda(x)b,\qquad
a=\frac{c\times d}{|c|^2},\qquad b=c,\qquad
\lambda(x)=\frac{c\cdot x}{|c|^2}.}
$$
Here $a\cdot b=0$ and $|b|=|c|$, as required. Conversely,
$$
(a+\lambda c)\times c
=\frac{(c\times d)\times c}{|c|^2}
=d-\frac{(c\cdot d)c}{|c|^2}=d.
$$
Thus the perpendicularity assumption is sufficient, and the solutions form a <straight line> parallel to $c$. When $d\ne0$, the <vector> $d$ is perpendicular both to $c$ and to every position <vector> on the line; it is the normal to the plane through the origin containing that line. The point $a=c\times d/|c|^2$ is the perpendicular foot from the origin. Its distance from the origin is $|d|/|c|$, since $c\cdot d=0$.
For a general parametric <straight line>, minimize the squared <Euclidean distance>
$$
|a+\lambda b-y|^2=|a-y|^2+2\lambda b\cdot(a-y)+\lambda^2|b|^2.
$$
This strictly convex <quadratic function> has its unique minimum at $\lambda=b\cdot(y-a)/|b|^2$, giving the <orthogonal projection>
$$
\boxed{x_{\rm nearest}=a+\frac{b\cdot(y-a)}{|b|^2}b.}
$$
For the <cross product> description, $a\cdot c=0$, so this becomes
$$
\boxed{x_{\rm nearest}=\frac{c\times d}{|c|^2}+\frac{c\cdot y}{|c|^2}c.}
$$
For the two planes, put $s=m\cdot n$ and $c=m\times n$. Nonparallel unit normals give $|s|<1$ and $|c|^2=1-s^2$. Seek the perpendicular foot in their span, $a=um+vn$. Its two <dot products> impose $u+sv=\mu$ and $su+v=\nu$. Solving gives both requested descriptions of the intersection:
$$
\boxed{x=\frac{(\mu-s\nu)m+(\nu-s\mu)n}{1-s^2}+\lambda(m\times n),\qquad\lambda\in\mathbb R,}
$$
and
$$
\boxed{x\times(m\times n)=\nu m-\mu n.}
$$
Indeed the <vector triple product identity> gives $x\times(m\times n)=m(x\cdot n)-n(x\cdot m)$. Conversely, since $m,n$ are linearly independent, equality to $\nu m-\mu n$ forces both plane equations. Thus neither description introduces extra points.
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