We need to find the smallest prime number $p$ that cannot be represented in the form $|3^a - 2^b|$, where $a$ and $b$ are non-negative integers.

First, we verify that all primes less than 41 can be expressed in the form $|3^a - 2^b|$:

- For $p = 2$: $2 = |3^0 - 2^1|$
- For $p = 3$: $3 = |3^1 - 2^0|$
- For $p = 5$: $5 = |3^1 - 2^2|$
- For $p = 7$: $7 = |3^2 - 2^3|$
- For $p = 11$: $11 = |3^2 - 2^5|$
- For $p = 13$: $13 = |3^3 - 2^3|$
- For $p = 17$: $17 = |3^3 - 2^4|$
- For $p = 19$: $19 = |3^3 - 2^5|$
- For $p = 23$: $23 = |3^3 - 2^6|$
- For $p = 29$: $29 = |3^3 - 2^7|$
- For $p = 31$: $31 = |3^4 - 2^5|$
- For $p = 37$: $37 = |3^4 - 2^6|$

Now, we check for $p = 41$:

### Case 1: $3^a - 2^b = 41$
- Since $3^a \equiv 0 \pmod{3}$, we have $-2^b \equiv 2 \pmod{3}$, implying $2^b \equiv 1 \pmod{3}$. This occurs when $b$ is even.
- Since $2^b \equiv 0 \pmod{4}$, we have $3^a \equiv 1 \pmod{4}$, implying $a$ is even.
- Let $a = 2j$ and $b = 2k$. Then $(3^j)^2 - (2^k)^2 = 41$, which factors as $(3^j - 2^k)(3^j + 2^k) = 41$.
- Since 41 is prime, $3^j - 2^k = 1$ and $3^j + 2^k = 41$. Adding these gives $3^j = 21$, which is not possible.

### Case 2: $2^b - 3^a = 41$
- Since $3^a \equiv 0 \pmod{3}$, we have $2^b \equiv 2 \pmod{3}$, implying $b$ is odd.
- Since $2^b \equiv 0 \pmod{4}$, we have $-3^a \equiv 1 \pmod{4}$, implying $3^a \equiv 3 \pmod{4}$, so $a$ is odd.
- Let $a = 2j + 1$ and $b = 2k + 1$. Then $2^b - 3^a \equiv 1 \pmod{8}$.
- Checking values, we find $2^b \equiv 4 \pmod{8}$, which is not possible since $b$ is odd.

Since both cases have no solutions, we conclude that 41 cannot be represented in the form $|3^a - 2^b|$.

The answer is $\boxed{41}$.