Step 1 :Let's denote the number of $8 tickets as x, $10 tickets as y, and $12 tickets as z.
Step 2 :From the problem, we know that the total number of tickets sold is 300, so we have the equation \(x + y + z = 300\).
Step 3 :The total income from ticket sales was $2680, so we have the equation \(8x + 10y + 12z = 2680\).
Step 4 :The combined number of $8 and $10 tickets sold was 5 times the number of $12 tickets sold, so we have the equation \(x + y = 5z\).
Step 5 :So, the correct system of linear equations is: \[\left\{\begin{array}{l}x+y+z=300 \\ 8 x+10 y+12 z=2680 \\ x+y=5z\end{array}\right.\]
Step 6 :Solving this system of equations, we find that \(x = 210\), \(y = 40\), and \(z = 50\).
Step 7 :Final Answer: There were \(\boxed{210}\) tickets of $8, \(\boxed{40}\) tickets of $10, and \(\boxed{50}\) tickets of $12 sold.