algebra precalculus - Find $cos(2alpha)$ given $cos(theta -alpha)$ and $sin(theta +alpha)$

My question is:

If $\cos(\theta -\alpha) = \frac{3}{5}$ and $\sin(\theta +\alpha) =\frac{12}{13}$, find $\cos(2\alpha)$.

Attempt I:
$$\begin{align*} &\cos^2(\theta -\alpha)+\sin^2(\theta +\alpha) = \frac{9}{25} + \frac{144}{169}\\ \Rightarrow &\cos^2(\theta -\alpha)- \cos^2(\theta +\alpha) = \frac{9}{25} + \frac{144}{169} - 1 = \frac{896}{4225}. \end{align*}$$
But I thought it won't work.

Then I tried this:

Attempt II:
$$\begin{align*} &\begin{cases} \cos\theta \cos\alpha + \sin\theta \sin\alpha = \cos(\theta -\alpha) = \frac{3}{5},\\ \sin\theta \cos\alpha + \cos\theta \sin\alpha = \sin(\theta +\alpha) =\frac{12}{13}, \end{cases}\\ &\cos\theta (\sin\alpha + \cos\alpha) + \sin\theta (\sin\alpha + \cos\alpha) = \frac{99}{65},\\ &\sqrt{2}\left[\sin\left(\alpha+\frac{\pi}{4}\right) \cos\left(\alpha -\frac{\pi}{4}\right)\right] = \frac{99}{65}. \end{align*}$$
But I thought here its better to convert both parts to cosines, so I did:
$$\begin{align*} &\sqrt{2}\left[\cos\left(\alpha-\frac{\pi}{4}\right) \cos\left(\alpha -\frac{\pi}{4}\right)\right] = \frac{99}{65},\\ &\sqrt{2} \cos^2\left(\alpha-\frac{\pi}{4}\right) = \frac{99}{65}. \end{align*}$$
But I think it also didn't work ....

Please guide. Thanks.