8. Given $a, b, c \in R, a^{2}+b^{2}+c^{2}=1$, then the minimum value of $3 a b-3 b c+2 c^{2}$ is $\qquad$

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
8. $-\frac{1+\sqrt{7}}{2}$

Analysis: Considering undetermined coefficients, by the AM-GM inequality, for any $m, n>0$, we have $3 a b-3 b c+2 c^{2} \geq-\frac{3}{2}\left(m \cdot a^{2}+\frac{1}{m} b^{2}\right)-\frac{3}{2}\left(n \cdot b^{2}+\frac{1}{n} c^{2}\right)+2 c^{2}$. We only need to let $-\frac{3}{2} m=-\frac{3}{2 m}-\frac{3}{2} n=2-\frac{3}{2 n}$, which gives $n=m-\frac{1}{m}$. Therefore, $(m+2)\left(3 m^{2}-2 m-2\right)=0$. Since $m, n>0$, we have $m=\frac{1+\sqrt{7}}{3}$, and thus $n=\frac{5-\sqrt{7}}{6}$. This way,
$3 a b-3 b c+2 c^{2} \geq-\frac{1+\sqrt{7}}{2}\left(a^{2}+b^{2}+c^{2}\right)=-\frac{1+\sqrt{7}}{2}$, when $b=-\frac{1+\sqrt{7}}{2} a, c=\frac{5-\sqrt{7}}{6} b$, the minimum value is achieved. Therefore, the minimum value of $3 a b-3 b c+2 c^{2}$ is $-\frac{1+\sqrt{7}}{2}$.