Solve for the missing number in the following equations: - (12 / 3) = 6 18 + + 12 = 98 5 + + 12 = 95

The problem is $\square - (12 \div 3) = 6$.
Let the missing number be $x$.

$$x - (12 \div 3) = 6$$

First, calculate the division in the parentheses:

$$x - 4 = 6$$

Add 4 to both sides of the equation:

$$x = 6 + 4$$ $$x = 10$$

The student's work incorrectly substituted 30 for the missing number, which led to $30 - 4 = 26$, not 6. The subsequent step $18 \div 3 = 6$ is a separate true statement but does not follow from the equation.
The missing number is $\boxed{10}$.

Let the missing number be $y$.

$$18 + y + 12 = 98$$

Combine the constant terms on the left side:

$$30 + y = 98$$

Subtract 30 from both sides of the equation:

$$y = 98 - 30$$ $$y = 68$$

The missing number is $\boxed{68}$.

Let the missing number be $z$.

$$5 + z + 12 = 95$$

Combine the constant terms on the left side:

$$17 + z = 95$$

Subtract 17 from both sides of the equation:

$$z = 95 - 17$$ $$z = 78$$

The missing number is $\boxed{78}$.