At a school with 100 students, 34 were taking Arabic, 34 Bulgarian, and 32 Chinese. 11 students take only Arabic, 11 of take only Bulgarian, and 12 take only Chinese. In addition, 15 are taking both Arabic and Bulgarian, some of whom also take Chinese. How many students are taking all three languages? None of these three languages? $$ \square $$ students are taking all three languages. students are taking none of the three languages.

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Let 
$$
\begin{aligned}
n(A) &= 34,\quad n(B)=34,\quad n(C)=32,\[1mm]
A_{\text{only}} &= 11,\quad B_{\text{only}}=11,\quad C_{\text{only}}=12,\[1mm]
n(A\cap B) &= 15 \quad \text{(includes those taking all three),}\[1mm]
x &= n(A\cap B\cap C) \quad \text{(number taking all three)}.
\end{aligned}
$$

The number taking only Arabic and Bulgarian (and not Chinese) is 
$$
( A\cap B)_{\text{only}} = 15 - x.
$$

Now, consider $$n(A)$$. Writing it as the sum of disjoint parts,
$$
n(A) = A_{\text{only}} + ( A\cap B)_{\text{only}} + ( A\cap C)_{\text{only}} + x.
$$
Substitute the known numbers:
$$
34 = 11 + (15-x) + ( A\cap C)_{\text{only}} + x.
$$
Simplify:
$$
34 = 11 + 15 + ( A\cap C)_{\text{only}} \quad \Longrightarrow \quad 34 = 26 + (A\cap C)_{\text{only}}.
$$
Thus,
$$
(A\cap C)_{\text{only}} = 34 - 26 = 8.
$$

Similarly, considering $$n(B)$$:
$$
n(B) = B_{\text{only}} + ( A\cap B)_{\text{only}} + ( B\cap C)_{\text{only}} + x.
$$
Substitute the numbers:
$$
34 = 11 + (15-x) + ( B\cap C)_{\text{only}} + x,
$$
which leads to:
$$
34 = 26 + (B\cap C)_{\text{only}} \quad \Longrightarrow \quad (B\cap C)_{\text{only}} = 8.
$$

Now, for $$n(C)$$:
$$
n(C) = C_{\text{only}} + ( A\cap C)_{\text{only}} + (B\cap C)_{\text{only}} + x.
$$
Substitute the known values:
$$
32 = 12 + 8 + 8 + x.
$$
Calculate:
$$
32 = 28 + x \quad \Longrightarrow \quad x = 32 - 28 = 4.
$$

Thus, the number of students taking all three languages is 
$$
\boxed{4}.
$$

Next, we find the number of students taking at least one language. This is the sum of the disjoint parts:
$$
\begin{aligned}
\text{Only Arabic} &: 11,\[1mm]
\text{Only Bulgarian} &: 11,\[1mm]
\text{Only Chinese} &: 12,\[1mm]
\text{Arabic and Bulgarian only} &: 15 - x = 15 - 4 = 11,\[1mm]
\text{Arabic and Chinese only} &: 8,\[1mm]
\text{Bulgarian and Chinese only} &: 8,\[1mm]
\text{All three} &: x = 4.
\end{aligned}
$$
The total is:
$$
11 + 11 + 12 + 11 + 8 + 8 + 4 = 65.
$$

Since there are 100 students in total, the number taking none of the three languages is:
$$
100 - 65 = 35.
$$

Thus, the final answers are:
$$
\boxed{4} \quad \text{students are taking all three languages,}
$$
$$
\boxed{35} \quad \text{students are taking none of the three languages.}
$$
4 students are taking all three languages, and 35 students are taking none of the three languages.

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