If $$X=\begin{bmatrix}x \\y \\z \end{bmatrix}$$ is a solution of the system of equations AX= B, where adj $$A= \begin{bmatrix}4 & 2 & 2 \\-5 & 0 & 5 \\1 & -2 & 3 \end{bmatrix}$$ and $$B=\begin{bmatrix}4 \\0 \\2 \end{bmatrix}$$, then |x+y+z| is equal to :

Given: $$\text{adj}(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$$ and $$B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$$.

We know that $$A \cdot \text{adj}(A) = |A| \cdot I$$. Also, $$A^{-1} = \frac{\text{adj}(A)}{|A|}$$.

First, find $$|A|$$. Since $$|\text{adj}(A)| = |A|^{n-1} = |A|^2$$ for a $$3 \times 3$$ matrix:

$$|\text{adj}(A)| = 4(0 \cdot 3 - 5 \cdot (-2)) - 2((-5)(3) - 5 \cdot 1) + 2((-5)(-2) - 0 \cdot 1)$$

$$= 4(0 + 10) - 2(-15 - 5) + 2(10 - 0) = 40 + 40 + 20 = 100$$

So $$|A|^2 = 100$$, giving $$|A| = 10$$ (taking positive value).

The solution of $$AX = B$$ is $$X = A^{-1}B = \frac{1}{|A|}\text{adj}(A) \cdot B$$.

$$\text{adj}(A) \cdot B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix}$$

$$X = \frac{1}{10}\begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$

So $$x = 2, y = -1, z = 1$$.

$$|x + y + z| = |2 - 1 + 1| = |2| = 2$$.

The correct answer is Option C: 2.