The motion of the airplane is described by the equations of kinematics for constant acceleration. The initial velocity is $v_0 = 60$ m/s and the acceleration is $a = -6$ m/s² (negative as it is a deceleration).

First, we determine the time required for the airplane to come to a complete stop ($v=0$). The velocity equation is:

$$v = v_0 + at$$

Setting $v=0$ gives the stopping time, $t_{stop}$:

$$0 = v_0 + at_{stop}$$ $$t_{stop} = -\frac{v_0}{a}$$

Substituting the values:

$$t_{stop} = -\frac{60 \text{ m/s}}{-6 \text{ m/s}^2} = 10 \text{ s}$$

The question asks for the position at $t = 12$ s. Since this is after the airplane has come to a stop at $t_{stop} = 10$ s, the airplane's position will be its final stopping distance.

The position $x$ is given by the kinematic equation:

$$x = v_0 t + \frac{1}{2}at^2$$

We evaluate this equation at the time of stopping, $t = t_{stop} = 10$ s, to find the stopping distance, $x_{stop}$.

$$x_{stop} = v_0 t_{stop} + \frac{1}{2}a t_{stop}^2$$

Substituting the values:

$$x_{stop} = (60 \text{ m/s})(10 \text{ s}) + \frac{1}{2}(-6 \text{ m/s}^2)(10 \text{ s})^2$$ $$x_{stop} = 600 \text{ m} - 3(100) \text{ m} = 300 \text{ m}$$

The position of the airplane at 12 seconds is this final stopping distance.