直线 $x - y = 3$ 与曲线 $x^2 - 3xy + y^2 + 19 = 0$ 交于点 $P$ 和 $Q$。
求 $P$ 和 $Q$ 的坐标。

1. Express $x$ in terms of $y$ using the linear equation

From $x - y = 3$, we have:
$x = y + 3$

2. Substitute the expression for $x$ into the quadratic equation

Substitute $x = y + 3$ into $x^2 - 3xy + y^2 + 19 = 0$:
$(y + 3)^2 - 3(y + 3)y + y^2 + 19 = 0$

3. Expand and simplify to form a quadratic equation in $y$

$y^2 + 6y + 9 - 3y^2 - 9y + y^2 + 19 = 0$
$-y^2 - 3y + 28 = 0$
$y^2 + 3y - 28 = 0$

4. Solve the quadratic equation for $y$

$(y + 7)(y - 4) = 0$
$y = -7 \text{ or } y = 4$

5. Find the corresponding values of $x$ and state the coordinates

When $y = -7$, $x = -7 + 3 = -4$
When $y = 4$, $x = 4 + 3 = 7$

The coordinates of $P$ and $Q$ are $(-4, -7)$ and $(7, 4)$.