Let M be a 3×33 \times 33×3 matrix such that M(100)=(123),M(010)=(012)M \begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}, M \begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}0\\1\\2\end{pmatrix}M100=123,M010=012 and M(001)=(−111)M \begin{pmatrix}0\\0\\1\end{pmatrix} = \begin{pmatrix}-1\\1\\1\end{pmatrix}M001=−111. If M(xyz)=(1711)M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}Mxyz=1711, then x+y+zx + y + zx+y+z equals :
4
5
7
11