1. Classify elements of $S$ by their remainder modulo $3$

Given $S=\{1,2,3,5,7,10,11\}.$

Reduce each element modulo $3$:

• $1\equiv 1 \pmod 3$
• $2\equiv 2 \pmod 3$
• $3\equiv 0 \pmod 3$
• $5\equiv 2 \pmod 3$
• $7\equiv 1 \pmod 3$
• $10\equiv 1 \pmod 3$
• $11\equiv 2 \pmod 3$

So the set splits as:

• residue $0$: $\{3\}$ (1 element)
• residue $1$: $\{1,7,10\}$ (3 elements)
• residue $2$: $\{2,5,11\}$ (3 elements)

2. Let us count subsets whose sum is divisible by $3$

Suppose a subset contains:

• $a$ elements from residue $1$ class,
• $b$ elements from residue $2$ class,
• and possibly the element $3$ from residue $0$ class.

Since including $3$ does not change the sum modulo $3$, the condition is $a+2b\equiv 0 \pmod 3.$ Equivalently, $a-b\equiv 0 \pmod 3 \quad\Rightarrow\quad a\equiv b \pmod 3.$

Here $a,b\in\{0,1,2,3\}$.

3. Find all valid pairs $(a,b)$

We need $a\equiv b\pmod 3$. Possible values are:

• $(0,0)$
• $(1,1)$
• $(2,2)$
• $(3,3)$
• $(0,3)$
• $(3,0)$

Now count subsets for each pair.

From the 3 residue-1 elements, choosing $a$ elements can be done in $\binom{3}{a}$ ways. From the 3 residue-2 elements, choosing $b$ elements can be done in $\binom{3}{b}$ ways. The residue-0 element $3$ may be either included or not included: $2$ choices.

Hence total count is $2\sum \binom{3}{a}\binom{3}{b}$ over the valid pairs.

So $N=2\left[\binom30\binom30+\binom31\binom31+\binom32\binom32+\binom33\binom33+\binom30\binom33+\binom33\binom30\right].$

Compute: $\binom30=1,\quad \binom31=3,\quad \binom32=3,\quad \binom33=1.$ Thus $N=2\left[1\cdot 1+3\cdot 3+3\cdot 3+1\cdot 1+1\cdot 1+1\cdot 1\right]$ $=2(1+9+9+1+1+1)=2\cdot 22=44.$

4. Exclude the empty subset

The above count includes the empty subset, corresponding to:

• choosing $a=0$, $b=0$, and not choosing $3$.

But the question asks for non-empty subsets.

Therefore, $44-1=43.$

5. Final answer

The number of non-empty subsets whose sum is a multiple of $3$ is $\boxed{43}.$