= Solution
The squared <Euclidean distance> along the <line> is
$$
|a+td|^2=|a|^2+2t(a\cdot d)+t^2|d|^2,\qquad d=b\times c.
$$
Its minimum occurs at $t=0$ exactly when its <derivative> there vanishes, so \b[$a\cdot(b\times c)=0$]. Since $b,c$ are independent, their <linear span> is the two-dimensional <orthogonal complement> of $b\times c$. Consequently
$$
\boxed{a=\alpha b+\beta c}
$$
for unique real $\alpha,\beta$. This is the <closest point on the intersection of two planes>; the positive quadratic coefficient $|d|^2$ proves that the stationary distance is a strict <global minimum>.
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