[Algebra]
Problem: Determine all real numbers $a, b$ and $c$ for which the graph of the function $y=\log _{a}(x+b)+c$ passes through the points $P(3,5), Q(5,4)$ and $R(11,3)$.

Solution: Since the graph passes through $(3,5),(5,4)$ and $(11,3)$, we can substitute these three points and obtain the following three equations:

$$
\begin{aligned}
& 5=\log _{a}(3+b)+c \\
& 4=\log _{a}(5+b)+c \\
& 3=\log _{a}(11+b)+c
\end{aligned}
$$

Subtracting the second equation from the first and the third equation from the second, we obtain:

$$
\begin{aligned}
& 1=\log _{a}(3+b)-\log _{a}(5+b) \\
& 1=\log _{a}(5+b)-\log _{a}(11+b)
\end{aligned}
$$

Equating right sides and manipulating, we obtain the following equivalent equations:

$$
\begin{aligned}
\log _{a}(5+b)-\log _{a}(11+b) & =\log _{a}(3+b)-\log _{a}(5+b) \\
2 \log _{a}(5+b) & =\log _{a}(3+b)+\log _{a}(11+b) \\
\log _{a}\left((5+b)^{2}\right) & =\log _{a}((3+b)(11+b)) \quad(\text { using log laws }) \\
(5+b)^{2} & =(3+b)(11+b) \quad \text { (raising both sides to the power of } a) \\
25+10 b+b^{2} & =33+14 b+b^{2} \quad \\
-8 & =4 b \\
b & =-2
\end{aligned}
$$

Since $b=-2$, the equation $1=\log _{a}(3+b)-\log _{a}(5+b)$ becomes $1=\log _{a} 1-\log _{a} 3$.

Since $\log _{a} 1=0$ for every admissible value of $a$, then $\log _{a} 3=-1$ which gives $a=3^{-1}=\frac{1}{3}$.

Finally, the equation $5=\log _{a}(3+b)+c$ becomes $5=\log _{1 / 3}(1)+c$ and so $c=5$.

Therefore, $a=\frac{1}{3}, b=-2$, and $c=5$, which gives $y=\log _{1 / 3}(x-2)+5$.

Checking:

- When $x=3$, we obtain $y=\log _{1 / 3}(3-2)+5=\log _{1 / 3} 1+5=0+5=5$.
- When $x=5$, we obtain $y=\log _{1 / 3}(5-2)+5=\log _{1 / 3} 3+5=-1+5=4$.
- When $x=11$, we obtain $y=\log _{1 / 3}(11-2)+5=\log _{1 / 3} 9+5=-2+5=3$.

Final answer: $\frac{1}{3},-2,5$