Answer
No possible linear combination.
Work Step by Step
Let $v_1$ be the first column vector and $v_2$ be the second one, $v_3$ be the third one.
Let $av_1+bv_2=v_3$. Then $a+2b=3$, $7a+8b=9$, $4a+5b=7$ are the equations we have to satisfy.
Thus $a=3-2b$, thus $7(3-2b)+8b=9\\21-14b+8b=9\\12=6b\\b=2$
Thus $a=3-2(2)=-1$
We have to check if this solution satisfies the third equation.
$4a+5b=4(-1)+5(2)-4+10=6\ne7$.
Thus, there is no possible linear combination.
