First simplify $\frac{8}{12}$ to $\frac{2}{3}$. Then the equation becomes $\frac{x}{3}=\frac{2}{3}$. Multiply both sides by $3$ to get $x=2$.

A point $(x,y)$ lies on the graph of $y=\frac{12}{x}$ if its coordinates satisfy the equation. For $(2,6)$, substituting $x=2$ gives $y=\frac{12}{2}=6$, which matches the point. So $(2,6)$ lies on the graph.

Use the table values to write two equations.

From $Q(2)=11$:

so

From $Q(5)=5$:

so

which gives

Subtract the first equation from the second:

so $a=3$.

Then

Now find $Q(3)$:

So the correct answer is $7$.

Factor the numerator:

For any real number $x$ with $x\ne 3$, the common factor $x-3$ cancels, so

However, the original expression is not defined at $x=3$ because the denominator is $0$. Therefore, the statement that must be true is that $E(x)=x+3$ for every real number $x$ except $x=3$.

The line passes through $(0,6)$ and $(8,0)$, so its slope is

Using the point $(0,6)$, the equation of the line is

Set $y=3$:

So