Example 2 Given $x, y, z \in \mathbf{R}$, and satisfying $x+y+z=2$, find the minimum value of $T=x^{2}-2 x+2 y^{2}+3 z^{2}$.
Parse $T+1=(x-1)^{2}+2 y^{2}+3 z^{2}$ as a "sum of squares," and then multiply it by another "sum of squares," which results in the form of the product of "sum of squares" on the left side of the Cauchy-Schwarz inequality. Thus, $\frac{11}{6}(T+1)$
$$\begin{array}{l}
=\left[1^{2}+\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(\frac{\sqrt{3}}{3}\right)^{2}\right]\left[(x-1)^{2}+2 y^{2}+3 z^{2}\right] \\
\geqslant[(x-1)+y+z]^{2}=1,
\end{array}$$

Equality holds if and only if $(x-1): 1=\sqrt{2} y: \frac{\sqrt{2}}{2}=\sqrt{3} z: \frac{\sqrt{3}}{3}$, which means $x=\frac{17}{11}, y=\frac{3}{11}, z=\frac{2}{11}$, hence $T_{\text {min }}=-\frac{5}{11}$.