<s>[INST] Find perpendicular distance from point to line in 3D? I have a Line going through points B and C; how do I find the perpendicular distance to A?
$$A= (4,2,1)$$
$$B= (1,0,1)$$
$$C = (1,2,0)$$

 [/INST] Another approach (a very low-tech solution). By the Pythagorean theorem we have 
$$ AB=\sqrt{13},\qquad AC=\sqrt{10},\qquad BC=\sqrt{5}\tag{1}$$
hence $ABC$ is an acute triangle (since $AB^2<AC^2+BC^2$ and so on) and by calling $H_A$ the projection of $A$ on $BC$, we also have
$$ H_A B^2 - H_A C^2 = AB^2 - AC^2 = 3 \tag{2}$$
Since $H_A B + H_A C = BC = \sqrt{5}$, from $(2)$ it follows that $H_A B-H_A C = \frac{3}{\sqrt{5}}$, so:
$$ H_A B = \frac{4\sqrt{5}}{5}\qquad H_A C = \frac{\sqrt{5}}{5}\tag{3}$$
and:
$$\boxed{\phantom{\sum_{j=1}^{n^2}} H_A = \frac{4}{5}C+\frac{1}{5}B = \color{red}{\left(\frac{5}{5};\frac{8}{5};\frac{1}{5}\right)},\qquad AH_A^2 = AC^2-H_A C^2 = \color{red}{\frac{49}{5}}\phantom{\sum_{j=1}^{n^2}}}\tag{4} $$
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