已知_(1)=((dfrac {1)(3),-dfrac (2)(3),-dfrac (2)(3))}^T, _(2)=((-dfrac {2)(3),dfrac (1)(3),-dfrac (2)(3))}^T _(3)=((-dfrac {2)(3),-dfrac (2)(3),dfrac (1)(3))}^x,则_(1)=((dfrac {1)(3),-dfrac (2)(3),-dfrac (2)(3))}^T, _(2)=((-dfrac {2)(3),dfrac (1)(3),-dfrac (2)(3))}^T _(3)=((-dfrac {2)(3),-dfrac (2)(3),dfrac (1)(3))}^x在此基下的坐标为__________.(A).(-3,-2,-1) (B).(3,2,1) (C).(1,2,3) D.(-1,-2,-3)

步骤 1：设向量$\alpha$在此基下的坐标
设$\alpha = k_1a_1 + k_2a_2 + k_3a_3$，其中$a_1, a_2, a_3$是给定的基向量，$\alpha$是向量$(1,2,3)^T$，$k_1, k_2, k_3$是待求的坐标。

步骤 2：建立方程组
根据向量$\alpha$的坐标表示，可以得到以下方程组：
$$
\begin{cases}
\dfrac{1}{3}k_1 - \dfrac{2}{3}k_2 - \dfrac{2}{3}k_3 = 1 \\
-\dfrac{2}{3}k_1 + \dfrac{1}{3}k_2 - \dfrac{2}{3}k_3 = 2 \\
-\dfrac{2}{3}k_1 - \dfrac{2}{3}k_2 + \dfrac{1}{3}k_3 = 3
\end{cases}
$$

步骤 3：解方程组
解上述方程组，得到：
$$
\begin{cases}
k_1 = -3 \\
k_2 = -2 \\
k_3 = -1
\end{cases}
$$