Triangle Problem

Suppose $k$ is a real number such that

$\cos A = 2k$ , $\cos B = 9k$ and $\cos C = 12k$ .

Also, $\sin A = \sqrt{1-4k^{2}}$ , $\sin B = \sqrt{1-81k^{2}}$ and $\sin C = \sqrt{1-144k^{2}}$ .

In triangle $ABC$ , we have $A+B = 180^{o}-C$

$\cos (A+B) = \cos (180^{o}-C)$

$\cos A \cos B - \sin A \sin B = -\cos C$

$(2k)(9k) - \sqrt{1-4k^{2}}\sqrt{1-81k^{2}} = -12k$

$\sqrt{(1-4k^{2})(1-81k^{2})} = 18k^{2}+12k$

By squaring and multiplying,

$324k^{4}-85k^{2}+1 = 324k^{4}+432k^{3}+144k^{2}$

$432k^{3}+229k^{2}-1 = 0$

$(16k-1)(27k^{2}+16k+1) = 0$

So $k = \frac{1}{16}$

(Can you prove that the quadratic factor gives 'wrong' values of $k$ ?)

Then $\sin A = \sqrt{1-\frac{4}{256}} = \frac{6\sqrt{7}}{16}$

$\sin B = \sqrt{1-\frac{81}{256}} = \frac{5\sqrt{7}}{16}$

$\sin C = \sqrt{1-\frac{144}{256}} = \frac{4\sqrt{7}}{16}$

$\sin A : \sin B : \sin C = 6:5:4$

Therefore, $p = 6$ , $q = 5$ and $r = 4$ ;

and $pqr-(p+q+r) = 120-15 = 105$ .

I'm sorry but in this case, I have no idea on how to solve this problem.. I used trial and error where I divided both sides by 15, 16, etc.. I found out that if I get the sum of the arc cosine of 2/16, 9/16 and 12/16, it sums up to 180 (which is the desired angles for the triangle.. By getting angles A, B and C and dividing their sines, it yielded the ratio 6:5:4.. Afterwhich, I solved for the asked answer and it resulted 105..