Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$

Show for yourself that if $A,B,C$ are the angles of a triangle then $$ \cos^2A + \cos^2B + \cos^2C = 1-2\cos A \cos B \cos C $$

This is not very difficult, use the fact that $A+B+C = 180$ along with double angle identities.

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Therefore, if each of those ratios equaled $k$ , we get $62k^2 = 1-84k^3$, which can be solved using Cardano's formula (or you can use IVT to just assert the existence of a root) to get you a root like $k \approx 0.117928$. From here, you get that such a triangle in fact exists, and roughly has angles $76.36,69.28$ and $34.36$.

Strating from @Teresa Lisbon's answer, the exact results are $$k=\frac{31}{126} \left(2 \cos \left(\frac{1}{3} \left(2 \pi n-\cos ^{-1}\left(-\frac{17884}{29791}\right)\right)\right)-1\right)\qquad (n=0,1,2)$$ and this gives angles (in degrees) $a=76.358$, $b=69.281$, $c=34.361$.

Using algebra, the problem is very simple since it reduces to the equation $$a+\cos ^{-1}\left(\frac{3}{2} \cos (a)\right)+\cos ^{-1}\left(\frac{7 }{2} \cos (a)\right)=\pi$$ which has only one solution.

Using series expansion around $a=\frac \pi 2$ gives $$0=\frac{\pi }{2}+6 \left(a-\frac{\pi }{2}\right)+\frac{55}{8} \left(a-\frac{\pi }{2}\right)^3+\frac{4627}{128} \left(a-\frac{\pi }{2}\right)^5+O\left(\left(a-\frac{\pi }{2}\right)^7\right)$$ and series reversion leads to $$a \sim\frac{5 \pi }{12}+\frac{55 \pi ^3}{82944}+\frac{89 \pi ^5}{10616832}\approx 1.33212$$ while the "exact" solution is $a=1.33270$.

Hint use may use the identity

$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$$