Let's do a quick draw to help us visualize the problem:
That's a generic parallelogram, to verify that it's a parallelogram we can see that
[tex]\begin{gathered} AB=CD \\ \\ BC=AD \end{gathered}[/tex]The opposite lengths are equal, then, let's do something similar here, let's say that
[tex]\vec{AB}=\vec{CD}[/tex]then
[tex]\begin{gathered} \vec{AB}=B-A=(4,b,-9)-(a,-6,2)=(4-a,b+6,-9-2) \\ \\ \vec{AB}=(4-a,b+6,-11) \end{gathered}[/tex]And the vector CD
[tex]\begin{gathered} \vec{CD}=D-C=(-2,-5,11)-(3,5,c)=(-2-3,-5-5,11-c) \\ \\ \vec{CD}=(-5,-10,11-c) \end{gathered}[/tex]Let's impose our condition
[tex]\begin{gathered} \begin{equation*} \vec{AB}=\vec{CD} \end{equation*} \\ \\ (4-a,b+6,-11)=(-5,-10,11-c) \\ \\ \end{gathered}[/tex]Then
[tex]\begin{gathered} 4-a=-5 \\ \\ b+6=-10 \\ \\ 11-c=-11 \end{gathered}[/tex]By solving that equations we get
[tex]\begin{gathered} a=9 \\ \\ b=-16 \\ \\ c=22 \end{gathered}[/tex]
