Here is the solution for the provided equations and their classifications:

First Equation:

$-3x + 12 - 9x + 8 = -3(4x - 7)$

Simplify each side:

$-12x + 20 = -12x + 21$

Simplify further:

$-12x + 20 = -12x + 21$

Cancel out $-12x$ from both sides:

$20 = 21$

This is a contradiction, so the first equation has no solution.

Second Equation:

$-2(3x - 10) + 8x = 7x + 20 + 7x$

Simplify each side:

$-6x + 20 + 8x = 14x + 20$

Combine like terms:

$2x + 20 = 14x + 20$

Subtract $20$ from both sides:

$2x = 14x$

Subtract $2x$ from both sides:

$0 = 12x$

Divide by $12$:

$x = 0$

The second equation has one solution: $x = 0$.

Third Equation:

$2 + 4x - 10 + 6x = 2(5x - 4)$

Simplify each side:

$-8 + 10x = 10x - 8$

Simplify further:

$10x - 8 = 10x - 8$

This is an identity (true for all values of $x$), so the third equation has infinitely many solutions.

Final Classification:

1. First Equation: No solution.
2. Second Equation: One solution ($x = 0$).
3. Third Equation: Infinitely many solutions.

Let me know if you want a detailed explanation of any step or further clarifications!

Related Questions:

1. What steps are necessary to determine if an equation has no solution, one solution, or infinitely many solutions?
2. How can you verify your solution for linear equations?
3. What role does simplifying each side play in solving equations?
4. Are there specific cases where contradictions arise in solving equations?
5. What does it mean mathematically when an equation has infinitely many solutions?

Tip: Always simplify each side of the equation fully before attempting to solve it. This reduces errors and clarifies the solution process.