Determine which vectors are solutions of the system.
\begin{align*} & \hphantom{+}3x-2y-5z = \hphantom{+}4 \\ & \hphantom{+}2x+4y-\hphantom{1}z = \hphantom{+\llap{$0$}}2 \\ & {-}4x-8y+9z = {-}6 \end{align*}
I $\langle 4,-1,2 \rangle$
Substitute I into the last equation which evaluates to $\displaystyle 10=-6$, a contradiction.
Hence, I is not a solution.
II $\langle 4,2,0 \rangle$
Substitute II into the first equation which evaluates to $\displaystyle 8=4$, a contradiction.
Hence, II is not a solution.
III $\langle -\frac{3}{2},\frac{3}{4},-2 \rangle$
Substitute III into the second equation which evaluates to $\displaystyle -18=-6$, a contradiction.
Hence, III is not a solution.
IV $\langle -1,-1,-1 \rangle$
Substitute IV into the second equation which evaluates to $\displaystyle 5=2$, a contradiction.
Hence, IV is not a solution.
Correct?
I have check your calculations and they are exact. You probably could have double checked yourself. Believe in yourself!