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(C) Let $f(x) = ||x - 2| - |3 - x||$. By the triangle inequality,$||x - 2| - |3 - x|| \leq |(x - 2) - (3 - x)| = |2x - 5|$. More precisely,the express

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$3$

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(C) Let $f(x) = ||x - 2| - |3 - x||$. By the triangle inequality,$||x - 2| - |3 - x|| \leq |(x - 2) - (3 - x)| = |2x - 5|$. More precisely,the expression $||x - 2| - |3 - x||$ represents the absolute difference of the distances of $x$ from $2$ and $3$. For $x For $2 \leq x \leq 3$,$f(x) = |(x - 2) - (3 - x)| = |2x - 5|$,which ranges from $1$ to $0$ and back to $1$. For $x > 3$,$f(x) = |(x - 2) - (x - 3)| = |1| = 1$. Thus,the range of $f(x)$ is $[0, 1]$. For the equation $f(x) = 2 - a$ to have a solution,we must have $0 \leq 2 - a \leq 1$. This implies $a - 2 \leq 0$ and $2 - a \leq 1$,so $a \leq 2$ and $a \geq 1$. The integral values of $a$ are $1$ and $2$. The sum of these values is $1 + 2 = 3$.

Find the sum of all integral values of $a$ such that the equation $||x - 2| - |3 - x|| = 2 - a$ has a solution.

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