Find all the solutions to
\[\sqrt{3x^2 - 8x + 1} + \sqrt{9x^2 - 24x - 8} = 3.\]Enter all the solutions, separated by commas.
We can write the given equation as
\[\sqrt{3x^2 - 8x + 1} + \sqrt{3(3x^2 - 8x + 1) - 11} = 3.\]Thus, let $y = 3x^2 - 8x + 1,$ so
\[\sqrt{y} + \sqrt{3y - 11} = 3.\]Then $\sqrt{3y - 11} = 3 - \sqrt{y}.$  Squaring both sides, we get
\[3y - 11 = 9 - 6 \sqrt{y} + y.\]Then $20 - 2y = 6 \sqrt{y},$ so $10 - y = 3 \sqrt{y}.$  Squaring both sides, we get
\[y^2 - 20y + 100 = 9y.\]Then $y^2 - 29y + 100 = 0,$ which factors as $(y - 4)(y - 25) = 0.$  Thus, $y = 4$ or $y = 25.$  But only $y = 4$ satisfies $10 - y = 3 \sqrt{y}.$

Then $3x^2 - 8x + 1 = 4,$ so
\[3x^2 - 8x - 3 = 0.\]This leads to the solutions $\boxed{3, -\frac{1}{3}}.$  We check that these solutions work.