Given that $$|x - 1| + |x - 2| + |x - 3| \geq 6$$.
Case 1: When x > 3
$$(x - 1) + (x - 2) + (x - 3) \geq 6$$
$$x \geq 4$$
Therefore, the value of x $$\in$$ [4, $$\infty$$)Â
Case 2: When 2 < x < 3
$$(x - 1) + (x - 2) - (x - 3) \geq 6$$
$$x \geq 6$$
Therefore, no possible value of x in this domain.
Case 3: When 1 < x < 2
$$(x - 1) - (x - 2) - (x - 3) \geq 6$$
$$x \leq -2$$
Therefore, no possible value of x in this domain.
Case 3: When x < 1
$$-(x - 1) - (x - 2) - (x - 3) \geq 6$$
$$x \leq 0$$
Therefore, the value of x $$\in$$ (-$$\infty$$, 0]
Therefore, the value of x that will satisfy this inequality: x $$\in$$ (-$$\infty$$, 0] $$\cup$$ [4, $$\infty$$).
Hence, option B is the correct answer.
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