Show that vec a=\-1,3,2\ , vec b=\2,-3,-4\ , vec c=\-3,12,6\…

Show that $\vec a=\{-1,3,2\}$, $\vec b=\{2,-3,-4\}$, $\vec c=\{-3,12,6\}$ are coplanar and find their linear dependence.

Step 1: Coplanarity

\[ [\vec a,\vec b,\vec c]=\begin{vmatrix}-1&3&2\\2&-3&-4\\-3&12&6\end{vmatrix}=-1(-18+48)-3(12-12)+2(24-9)=-30+0+30=0. \]

Step 2: Linear dependence

From $\alpha\vec a+\beta\vec b+\gamma\vec c=\vec 0$ we get three equations. The third is twice the first, so two are independent: \[-\alpha+2\beta-3\gamma=0,\qquad 3\alpha-3\beta+12\gamma=0\ \Longrightarrow\ \beta=-\gamma,\quad \alpha=-5\gamma.\]

Taking $\gamma=1$ gives $-5\vec a-\vec b+\vec c=\vec 0$: $\vec c=5\vec a+\vec b$.

Check: $5\{-1,3,2\}+\{2,-3,-4\}=\{-3,12,6\}$ $\checkmark$