= Solution
With $b\cdot c=0$, substitute $a=\alpha b+\beta c$ into the <plane> constraints. They become $\alpha|b|^2=\lambda$ and $\beta|c|^2=\mu$, giving
$$
\boxed{\alpha=\frac{\lambda}{|b|^2},\qquad\beta=\frac{\mu}{|c|^2}.}
$$
The <Pythagorean theorem> for the two <orthogonal> terms in $a$ then gives the closest distance
$$
\boxed{d_{\min}=|a|=\sqrt{\frac{\lambda^2}{|b|^2}+\frac{\mu^2}{|c|^2}}.}
$$
Zero values of either <scalar> are allowed, including zero distance if both planes contain the origin.
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