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$ABCD$ is a tetrahedron with position vectors as $A(\hat{j}+2\hat{k})$, $B(3\hat{i}+\hat{k})$, $C(4\hat{i}+3\hat{j}+6\hat{k})$,$D(2\hat{i}+3\hat{j}+2\hat{k})$. We have to find the perpendicular distance between point $A$ and $BC$.

I can solve it by using coordinate geometry (not by vectors).

I want to know how can we solve it using vectors .

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Here's a hint: assume you already have the normal vector to the plane. Can you say anything about the perpendicular distance now? You should be able to take it from here.

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    Sorry , I am not getting after that – 2017-01-21
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    Assume you have already found the normal to the plane ABC, call it v. The cross product of v and BC after normalizing gives you a vector perpendicular to BC and also in the plane of ABC, call it n. So now you need to find *how far along n does A lie*? – 2017-01-21
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    How 'far' from where ? – 2017-01-21
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    Any point on the line AB will work. Take B to be the point, and calculate the dot product between n and (A - B). That will give you the distance you want. – 2017-01-21