If $\vec a, \vec b, \vec c$ are position vectors of three non-collinear points A, B, C respectively, the shortest distance of A from BC is

$\sqrt{|\vec b-\vec a|^2-\left(\frac{(\vec a-\vec b)(\vec c-\vec b)}{|\vec c-\vec b|}\right)^2}$

The vector equation of the line BC is

$\vec r=\vec b+λ(\vec c-\vec b)$

Clearly,

BM = Projection of $\vec{BA}$ on $\vec{BC}$

$⇒BM =\frac{\vec{BA}.\vec{BC}}{|\vec{BC}|}=\frac{(\vec a-\vec b).(\vec c-\vec b)}{|\vec c-\vec b|}$

∴ Required distance = $AM=\sqrt{AB^2-BM^2}$

⇒ Required distance = $\sqrt{|\vec b-\vec a|^2-\left(\frac{(\vec a-\vec b)(\vec c-\vec b)}{|\vec c-\vec b|}\right)^2}$