23. Solution: (1) Since $[(x-1) + (y+1) + (z+1)]^2$
$= (x-1)^2 + (y+1)^2 + (z+1)^2 + 2[(x-1)(y+1) + (y+1)(z+1) + (z+1)(x-1)]$
$\leq 3[(x-1)^2 + (y+1)^2 + (z+1)^2]$,
from the given condition we have $(x-1)^2 + (y+1)^2 + (z+1)^2 \geq \frac{4}{3}$,
with equality if and only if $x = \frac{5}{3}$, $y = -\frac{1}{3}$, $z = -\frac{1}{3}$.
Therefore, the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$ is $\frac{4}{3}$.
(2) Since
$[(x-2) + (y-1) + (z-a)]^2$
$= (x-2)^2 + (y-1)^2 + (z-a)^2 + 2[(x-2)(y-1) + (y-1)(z-a) + (z-a)(x-2)]$
$\leq 3[(x-2)^2 + (y-1)^2 + (z-a)^2]$,
from the given condition $(x-2)^2 + (y-1)^2 + (z-a)^2 \geq \frac{(2+a)^2}{3}$,
with equality if and only if $x = \frac{4-a}{3}$, $y = \frac{1-a}{3}$, $z = \frac{2a-2}{3}$.
Therefore, the minimum value of $(x-2)^2 + (y-1)^2 + (z-a)^2$ is $\frac{(2+a)^2}{3}$.
From the given condition, $\frac{(2+a)^2}{3} \geq \frac{1}{3}$, solving gives $a \leq -3$ or $a \geq -1$.