= Solution
The first coordinate of the <vector triple product> is
$$
[a\times(b\times c)]_1=a_2(b_1c_2-b_2c_1)-a_3(b_3c_1-b_1c_3)
=b_1(a\cdot c)-c_1(a\cdot b).
$$
The other two coordinates follow by cyclically permuting the indices, proving the identity.
A point on the line has the form $r=b+s m$. Substituting in the <plane> equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne0$ the parameter, and therefore the intersection point, is unique. The <vector triple product> gives $n\times(b\times m)=(n\cdot m)b-(n\cdot b)m$, so \b[the unique intersection is]
$$
\boxed{r=b+\frac{(a-b)\cdot n}{m\cdot n}m
=\frac{(a\cdot n)m+n\times(b\times m)}{m\cdot n}.}
$$
For the parallel case $m\cdot n=0$, the line is entirely in the <plane> if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and <plane> require nonzero direction and normal vectors. This is the <line-plane intersection criterion>.
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