If $x$ is a real number such that $\max(\min(x, 2 - x), x - 4, 2x - 8) = \pi - 3$, then the number of possible values of $x$ is

Let $T = \pi - 3 \approx 0.14$. The equation is $\max(\min(x, 2-x),\ x-4,\ 2x-8) = T$.

For the maximum to equal $T$, at least one of the three quantities must equal $T$, and none can exceed $T$.

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Case 1: $\min(x, 2-x) = T$.

If $x \leq 1$, $\min = x = T$, so $x = T \approx 0.14$.

If $x \geq 1$, $\min = 2-x = T$, so $x = 2 - T \approx 1.86$.

For each, check the other two are $\leq T$:

At $x = T$: $x - 4 \approx -3.86$, $2x - 8 \approx -7.72$. Both $\leq T$. Valid.

At $x = 2 - T$: $x - 4 \approx -2.14$, $2x - 8 \approx -4.28$. Both $\leq T$. Valid.

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Case 2: $2x - 8 = T$, so $x = \dfrac{T + 8}{2} \approx 4.07$.

Check others $\leq T$:

$\min(x, 2-x) = 2 - x \approx -2.07 \leq T$. Valid.

$x - 4 = \dfrac{T + 8}{2} - 4 = \dfrac{T}{2} \approx 0.07 \leq T$. Valid.

Valid solution.

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Case 3: $x - 4 = T$, so $x = T + 4 \approx 4.14$.

Check $2x - 8 = 2T + 8 - 8 = 2T \approx 0.28$. But $2T > T$, contradicting the maximum being $T$. Invalid.

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Total valid values of $x$: two from Case $1$ and one from Case $2$.

Answer $= 3$