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(B) Since we know the action of $M$ on the standard basis vectors, the columns of $M$ are the images of the standard basis vectors. Thus, $M = \begin{

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$5$

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(B) Since we know the action of $M$ on the standard basis vectors, the columns of $M$ are the images of the standard basis vectors. Thus, $M = \begin{pmatrix} 1 & 0 & -1 \ 2 & 1 & 1 \ 3 & 2 & 1 \end{pmatrix}$.We need to solve the system of linear equations $M \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 7 \ 11 \end{pmatrix}$.This gives the equations:$1) x - z = 1 \Rightarrow x = z + 1$$2) 2x + y + z = 7$$3) 3x + 2y + z = 11$Substituting $x = z + 1$ into equations $(2)$ and $(3)$:$2(z + 1) + y + z = 7 \Rightarrow 3z + y = 5 \Rightarrow y = 5 - 3z$$3(z + 1) + 2(5 - 3z) + z = 11 \Rightarrow 3z + 3 + 10 - 6z + z = 11 \Rightarrow -2z = -2 \Rightarrow z = 1$Now, find $x$ and $y$:$x = 1 + 1 = 2$$y = 5 - 3(1) = 2$Finally, $x + y + z = 2 + 2 + 1 = 5$.

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