已知: alpha_1, alpha_2, alpha_3 线性无关，beta_1 = 2alpha_2 - alpha_3，beta_2 = -alpha_1 + 2alpha_2，beta_3 = alpha_1 - alpha_2 + 3alpha_3 证明: beta_1, beta_2, beta_3 线性无关。

设 $k_1 \beta_1 + k_2 \beta_2 + k_3 \beta_3 = 0$，代入已知条件得
$k_1(2a_2 - a_3) + k_2(-a_1 + 2a_2) + k_3(a_1 - a_2 + 3a_3) = 0.$
合并同类项得
$(-k_2 + k_3)a_1 + (2k_1 + 2k_2 - k_3)a_2 + (-k_1 + 3k_3)a_3 = 0.$
由 $a_1, a_2, a_3$ 线性无关，系数必须全为零，即
$\begin{cases}-k_2 + k_3 = 0, \\2k_1 + 2k_2 - k_3 = 0, \\-k_1 + 3k_3 = 0.\end{cases}$
解得 $k_1 = k_2 = k_3 = 0$。
结论： $\beta_1, \beta_2, \beta_3$ 线性无关。

$\boxed{\beta_1, \beta_2, \beta_3 \text{ 线性无关}}$