To solve this problem, we are looking for the smallest positive integer $N$ that satisfies the following conditions:

1. $N \equiv 2 \pmod{3}$
2. $N \equiv 2 \pmod{5}$
3. $N \equiv 2 \pmod{7}$

Each of these conditions states that if the pencils are divided into rows of 3, 5, or 7, there are always 2 pencils left over. This means $N - 2$ must be divisible by 3, 5, and 7.

Step 1: Rewrite the Problem

Since $N \equiv 2 \pmod{3}$, $N \equiv 2 \pmod{5}$, and $N \equiv 2 \pmod{7}$, we can rewrite these congruences as: $N = 3k + 2$ $N = 5m + 2$ $N = 7n + 2$ for integers $k$, $m$, and $n$.

Step 2: Reduce to a Common Congruence

Since $N - 2$ must be divisible by 3, 5, and 7, the smallest such number $N - 2$ that meets these criteria is the least common multiple (LCM) of 3, 5, and 7.

The LCM of 3, 5, and 7 is: [ \text{LCM}(3, 5