Let $$M$$ be a $$3 \times 3$$ matrix such that $$M\begin{bmatrix}1\\0\\0\end{bmatrix} = \begin{bmatrix}1\\2\\3\end{bmatrix}$$, $$M\begin{bmatrix}0\\1\\0\end{bmatrix} = \begin{bmatrix}0\\1\\0\end{bmatrix}$$, $$M\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}-1\\1\\1\end{bmatrix}$$. If $$M\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}1\\7\\11\end{bmatrix}$$, then $$x + y + z$$ is equal to :

The images of the standard basis vectors give the columns of the matrix $$M$$.

First column (image of $$[1,0,0]^T$$): $$\begin{bmatrix}1\\2\\3\end{bmatrix}$$.
Second column (image of $$[0,1,0]^T$$): $$\begin{bmatrix}0\\1\\0\end{bmatrix}$$.
Third column (image of $$[0,0,1]^T$$): $$\begin{bmatrix}-1\\1\\1\end{bmatrix}$$.

Hence

$$ M= \begin{bmatrix} 1 & 0 & -1\\ 2 & 1 & 1\\ 3 & 0 & 1 \end{bmatrix}. $$

We are told $$M\begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}1\\7\\11\end{bmatrix}$$, so we must solve

$$ \begin{bmatrix} 1 & 0 & -1\\ 2 & 1 & 1\\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}1\\7\\11\end{bmatrix}. $$

This gives three linear equations:

$$x - z = 1 \; -(1)$$
$$2x + y + z = 7 \; -(2)$$
$$3x + z = 11 \; -(3)$$

From $$(1)$$: $$x = 1 + z$$.

Substitute into $$(3)$$:
$$3(1+z) + z = 11 \Longrightarrow 3 + 4z = 11 \Longrightarrow 4z = 8 \Longrightarrow z = 2.$$

Then $$x = 1 + z = 1 + 2 = 3$$.

Substitute $$x=3,\, z=2$$ in $$(2)$$:
$$2(3) + y + 2 = 7 \Longrightarrow 6 + y + 2 = 7 \Longrightarrow y = -1.$$

Finally,

$$x + y + z = 3 + (-1) + 2 = 4.$$

Option A which is: $$4$$