Given:
[tex]\begin{cases}\mathbf a+\mathbf b+\mathbf c=\mathbf0&(1)\\\|\mathbf a\|=1&(2)\\\|\mathbf b\|=2&(3)\\\|\mathbf c\|=3&(4)\end{cases}[/tex]
Take the dot product of both sides of (1) with [tex]\mathbf a[/tex]. You end up with
[tex]\mathbf a\cdot(\mathbf a+\mathbf b+\mathbf c)=\mathbf a\cdot\mathbf0[/tex]
[tex]\mathbf a\cdot\mathbf a+\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c=0[/tex]
[tex]\|\mathbf a\|^2+\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c=0[/tex]
[tex]\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c=-1[/tex]
Doing the same thing with [tex]\mathbf b[/tex] and [tex]\mathbf c[/tex], you end up with the system
[tex]\begin{cases}\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c=-1\\\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf c=-4\\\mathbf a\cdot\mathbf c+\mathbf b\cdot\mathbf c=-9\end{cases}\implies \begin{cases}\mathbf a\cdot\mathbf b=2\\\mathbf a\cdot\mathbf c=-3\\\mathbf b\cdot\mathbf c=-6\end{cases}[/tex]
and so
[tex]\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c+\mathbf b\cdot\mathbf c=2-3-6=-7[/tex]
