\[\boxed{\text{646\ (646).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x^{2} - ax + a - 3 = 0\]
\[по\ теореме\ Виета:x_{1} + x_{2} = a;\]
\[\text{\ \ }x_{1}x_{2} = a - 3\]
\[\left( x_{1} + x_{2} \right)^{2} = (a)^{2}\]
\[x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2} = a^{2}\]
\[x_{1}^{2} + x_{2}^{2} + 2 \cdot (a - 3) = a^{2}\]
\[x_{1}^{2} + x_{2}^{2} = a^{2} - 2a + 6 =\]
\[= a^{2} - 2a + 1 + 5 =\]
\[= (a - 1)^{2} + 5,\]
\[то\ есть\ x_{1}^{2} + x_{2}^{2} - принимает\ \]
\[наименьшее\ значение\ при\ \]
\[a = 1\]
\[x_{1}^{2} + x_{2}^{2} = 1^{2} - 2 \cdot 1 + 6 =\]
\[= 1 - 2 + 6 = 5\]
\[Ответ:x_{1}^{2} + x_{2}^{2} = 5\ при\ a = 1\text{.\ \ }\]