Solution:

Consider the polynomial $Q(x) = P(x) - 3$. Then $Q(x)$ has zeros and maximum value $0$ at $x=2, 3$. These conditions imply that $Q(x)$ has the form
$$Q(x) = A(x-2)^2(x-3)^2$$
That is, its graph looks like

because the values of $Q(x)$ should grow larger and larger through negative values as the variable $x$ goes to larger and larger values of both signs and the fact that the number of turning points should not exceed $4-1=3$ but should be more than $2$ (given by the maximum points).

Thus, $Q(1) = P(1) - 3 = 0 - 3 = -3$ implies
$$Q(1) = A(1-2)^2(1-3)^2 = A(1)^2(4) = 4A = -3$$
So $A = -\frac{3}{4}$.

Therefore,
$$Q(x) = -\frac{3}{4}(x-2)^2(x-3)^2$$
So
$$P(x) = Q(x) + 3 = -\frac{3}{4}(x-2)^2(x-3)^2 + 3$$
Now, compute $P(5)$:

First, $(5-2)^2 = 9$, $(5-3)^2 = 4$.
So
$$P(5) = -\frac{3}{4} \cdot 9 \cdot 4 + 3 = -\frac{3}{4} \cdot 36 + 3 = -27 + 3 = -24$$

Therefore, $P(5) = -24$.