Finding a vector perpendicular to a given vector in 3 dimensional space

Question

So if I have points $A(2, 6,-1)$ and $B(-3, 2, 8)$ that create vector $\vec{AB}$. I want to create perpendicular vector $\vec{AC}$, which has the same starting point $A$ and then unknown end point $C(x, y, z)$. If I also knew that $\vec{AC}$ has magnitude $3$, how do I then determine the point $C$?

Answer 1

Let $\vec{AC}(a,b,c)$.

Thus, $a^2+b^2+c^2=9$ and since $$\vec{AB}(-5,-4,9),$$ we obtain: $$-5a-4b+9c=0$$ and we got infinitely many these vectors and these points $C$.

If we want that $AC||DE$, where $\vec{DE}(14,5,3)$ then $a=14k$,$b=5k$, $c=3k$ and you can check if this possible.