A question about analytic geometry (vectors)

Question

Let $a , b, c$ be three vectors such that $a+3b+c = o$ where $o$ is $(0,0,0)$ vector. we also know $|a| = 3 , |b| = 4 , |c| =6$. we want to find $a.b + a.c + b.c$

I know we can solve it by saying $a+b+c = -2b$ and then squaring the sides. and at last we get the answer 3/2 . But my problem is that is it even possible for $|a| = 3 , |b| = 4 , |c| =6$ to be such that $a+3b+c = o$ ? I think it is not possible because of triangle inequality. If it is possible, how? and If it is not, why do we get an answer for it?

Answer 1

Dot the equation with each of the vectors \begin{eqnarray*} \mid a \mid^2 +3 a \cdot b +a \cdot c =0 \\ a \cdot b +3 \mid b \mid^2 +b \cdot c =0 \\ a \cdot c +3 b \cdot c+\mid c \mid^2 =0 \\ \end{eqnarray*} Now add the first and third equations and subtract the second $2(a \cdot b +a \cdot c +b \cdot c)+9+36-48=0$ ...

Answer 2

As Lord Shark of the Unknown pointed out in the comments, it's not possible to have three such vectors sum up to zero. But nevertheless the calculation is possible: it is just a series of logical passages which do not yet lead to a contradiction, but surely enough if you continued to use these data in other you might end up with some absurdity.

It's like saying, "let's assume that there exists a number $x$ such that $|x|>1$ and $2x=0$." You still can deduce from the first piece of data that $|2x|>2$ or that $|x|+3>4$...