<s>[INST] Math subject GRE Exam 9768 Q.16 The question is in the following picture:



And the answer is (D).

At first I agreed with this answer when I added the given three vectors component wise,but when I added the forth component in the three vectors I understood that it is impossible that we can get 5, so my opinion changed to the choice (A), so what is the misconception that I have or where is my wrong thinking, could anyone help me please?    
 [/INST] The question is the same as asking for what $m$ the following equation has a solution:
$$
\begin{pmatrix}
0&0&1\\
1&0&1\\
1&0&2\\
1&1&0
\end{pmatrix}
\begin{pmatrix}
x\\y\\z
\end{pmatrix}=
\begin{pmatrix}
1\\2\\m\\5
\end{pmatrix}\tag{1}
$$
Working on column vectors might make some observation easier. Note that (1) is equivalent to 
$$
x\begin{pmatrix}
0\\1\\1\\1
\end{pmatrix}
+y\begin{pmatrix}
0\\0\\0\\1
\end{pmatrix}
+z\begin{pmatrix}
1\\1\\2\\0
\end{pmatrix}=\begin{pmatrix}
1\\2\\m\\5
\end{pmatrix}\tag{2}
$$
It is very easy to observe that when $m=3$, (2) has a solution: $x=z=1$, $y=4$. This rules out A,B,C. To see $m=3$ is necessarily true, note that (2) implies:
$$
0x+0y+1z=1\\
1x+0y+1z=2\\
1x+0y+2z=m
$$
To make the linear system consistent so that it has a solution, one must have $1+2=m$ by observing that the coefficients of the first two rows add up to those of the third row. 
</s>