Conic sections revisted

Let the "height" of $P$ above (or below) the $xy-$ plane be $h$ , and let $\mathcal{l}$ be the line given by $y = mx+c, \space z = h \text{ where } c,h,m \in \mathbb{R}.$

Let the projection of a unit circle in the $xy-$ plane onto $P$ be the closed curve $C$ . The unit circle is given by $\left(x-a\right)^2 + \left(y-b\right)^2=1, \ z = 0$ , where $a,b \in \mathbb{R}$ . This can be parametrized to get: $x=a+\cos{t}, \quad y = b+\sin{t}, \quad z=0, \ 0 \le t \le 2\pi.$

The ellipse $C$ is then given by: $x=a+\cos{t}, \quad y=b+\sin{t}, \quad z=h+d\sin{t}, \ 0 \le t \le 2\pi.$

In the parametrization above, $d \in \mathbb{R}$ . Since $C$ is the projection of the unit circle in $z=0$ onto $P$ , the length of the minor axis of $C$ is invariant under rotation of $P$ (the semi-minor axes have length $1$ and lie on $\mathcal{l}$ ). So the region enclosed by $C$ having an area of $12\pi$ implies that the semi-major axes of the ellipse must have lengths $12$ units.

Therefore, we can write that $C$ is given by $x=a+\cos{t}, \space \ \space \ y = b+\sin{t}, \space \ \space \ z = h+\sqrt{12^2-1^2}\sin{t} \ \space \left(\text{as } d=\sqrt{12^2-1^2}\right)\\ \implies x=a+\cos{t}, \space \ \space \ y = b+\sin{t}, \space \ \space \ z = h+\sqrt{143}\sin{t}.$

Now we see that $C$ lies in the plane $P: \space z=\sqrt{143}y+h$ , which has direction vector $\displaystyle \left< 0, -\sqrt{143}, 1 \right>$ . The $xy-$ plane has direction vector $\left< 0,0,1 \right>$ so for the acute angle between these planes, we use the dot product $1=\sqrt{144}\cos{\theta} \implies \theta = \arccos{\left(\dfrac{1}{12}\right)} \approx 85.22^{\circ}.$

We know that $P$ is rotating about $\mathcal{l}$ at a rate of $1^{\circ}$ per second, so it takes $\boxed{85.22}$ seconds before $C$ encloses an area of $12\pi$ square units.