Answer
See below
Work Step by Step
Given $v_1=(1,2,3)\\
v_2=(4,5,6)\\
v_3=(7,8,9)$
We can notice that $$-v_1+2v_2=(-1+2,-2+10,-3+12)=(7,8,9)=v_3$$
Hence, the set $\{v_1,v_2,v_3\}$ is a linearly independent set in $R^3$
Let $a,b$ and $c$ be scalars
Assume that $v=av_1+bv_2+cv_3\in span \{v_1,v_2,v_3\}$
Obtain $av_1+bv_2+cv_3\\
=av_1+bv_2+c(-v_1+2v_2)\\
=av_1+bv_2-cv_1+2cv_2\\
=(a-c)v_1+(b+2c)v_2\\
\rightarrow v \in span \{v_1,v_2\}$
then we have $span\{v_1,v_2\} \subset span \{v_1,v_2,v_3\}\\
\rightarrow span \{v_1,v_2\}=span \{v_1,v_2,v_3\}$
Thus, $v_1$ and $v_2$ are not proportional
$\{v_1,v_2\}$ is a linearly independent set.
Geometrically, set $\{v_1,v_2,v_3\}$ is a plane through the origin determined by the vectors $v_1$ and $v_2$
