3. Let $N$ sq.m. need to be painted, $x$ - the number of students, $v-$ the productivity of each student, $t_{1}$ - the time it takes to complete the work. Then it follows that $x v t_{1}=N$ and each student should paint $v t_{1}=\frac{N}{x}$ sq. m.

In the condition of the problem, the master's productivity is greater than the productivity of each student by $k$ times $(k>1)$. Therefore, in the case where one of the students falls ill and is replaced by the master, it will be true that $((x-1) v+k v) t_{2}=N$, where $t_{2}$ - the time it takes to complete the work. Then $v t_{2}=\frac{N}{x+k-1}$ sq. m.

The condition that each remaining student painted $a$ sq.m. less is equivalent to the equation $\frac{N}{x}=\frac{N}{x+k-1}+a \Leftrightarrow(x+k-1) x=\frac{(k-1) N}{a}$. For $N=360, k=3, a=6$ we get $(x+2) x=120 \Rightarrow x=10$.
Answer: 9 .

Answer to the option: $7-2: 11$.