Variation 1. (Guangzhou, 2009 $\in \boldsymbol{R}$, and $a+b+c=2, a^{2}+2 b^{2}+3 c^{2}=4$, then the range of values for $a$ is $\qquad$ .
According to the problem, we have: $b+c=2-a, 2 b^{2}+3 c^{2}=4-$ $a^{2}$. By the Cauchy-Schwarz inequality, we get: $4-a^{2}=2 b^{2}+3 c^{2}=\frac{b^{2}}{1 / 2}+\frac{c^{2}}{1 / 3}$ $\geqslant \frac{(b+c)^{2}}{1 / 2+1 / 3}=\frac{6}{5}(2-a)^{2}$, so $5\left(4-a^{2}\right) \geqslant 6(2-$ $a)^{2}$,

Simplifying, we get: $11 a^{2}-24 a+4 \leqslant 0$, solving this yields: $2 / 11 \leqslant a \leqslant$ 2, so the range of values for $a$ is $[2 / 11,2]$.