Vector equation of the line $BC$ is $\overrightarrow r=\overrightarrow b+\lambda(\overrightarrow c-\overrightarrow b)$

Let $P$ be the foot of $\perp$ of the point $A$ on $BC$.

Then the required distance is $AP$

Complete the rt. angled $\Delta\:ABP$

Then using pythogorous theorem, $AP=\sqrt {AB^2-BP^2}$

$AB^2=|\overrightarrow b-\overrightarrow a|^2$

$BP$ is projection of the vector $\overrightarrow {BA}$ on the line $\overrightarrow {BC}$

$i.e.,$ $BP=\large\frac{(\overrightarrow a-\overrightarrow b).(\overrightarrow c-\overrightarrow b)}{|\overrightarrow c-\overrightarrow b|}$

$\therefore$ The required distance $AP=\sqrt {|\overrightarrow b-\overrightarrow a|^2-\bigg(\large\frac{(\overrightarrow a-\overrightarrow b).(\overrightarrow c-\overrightarrow b)}{|\overrightarrow c-\overrightarrow b|}\bigg)^2}$