Find with proof all positive $3$ digit integers $\overline{abc}$ satisfying 
\[ b\cdot \overline{ac}=c \cdot \overline{ab} +10 \]
To find all positive 3-digit integers $\overline{abc}$ that satisfy the equation 
\[ b \cdot \overline{ac} = c \cdot \overline{ab} + 10, \]
we will start by expressing $\overline{abc}$, $\overline{ac}$, and $\overline{ab}$ in terms of $a$, $b$, and $c$.

1. **Express the numbers in terms of $a$, $b$, and $c$:**
   - $\overline{abc} = 100a + 10b + c$
   - $\overline{ac} = 10a + c$
   - $\overline{ab} = 10a + b$

2. **Substitute these expressions into the given equation:**
   \[ b \cdot (10a + c) = c \cdot (10a + b) + 10 \]

3. **Expand and simplify the equation:**
   \[ 10ab + bc = 10ac + bc + 10 \]

4. **Subtract $bc$ from both sides:**
   \[ 10ab = 10ac + 10 \]

5. **Divide both sides by 10:**
   \[ ab = ac + 1 \]

6. **Rearrange the equation to isolate $a$:**
   \[ a(b - c) = 1 \]

7. **Analyze the equation $a(b - c) = 1$:**
   - Since $a$, $b$, and $c$ are digits (1 through 9), the only way for $a(b - c) = 1$ to hold is if $a = 1$ and $b - c = 1$ (since $a$ must be a positive integer and $b - c$ must also be an integer).

8. **Substitute $a = 1$ into the equation:**
   \[ 1(b - c) = 1 \]
   \[ b - c = 1 \]

9. **Solve for $b$ and $c$:**
   - $b = c + 1$

10. **Determine the valid values for $b$ and $c$:**
    - Since $b$ and $c$ are digits, $1 \leq c \leq 8$ and $b = c + 1$.

11. **List all possible values of $\overline{abc}$:**
    - For $c = 1$, $b = 2$: $\overline{abc} = 112$
    - For $c = 2$, $b = 3$: $\overline{abc} = 123$
    - For $c = 3$, $b = 4$: $\overline{abc} = 134$
    - For $c = 4$, $b = 5$: $\overline{abc} = 145$
    - For $c = 5$, $b = 6$: $\overline{abc} = 156$
    - For $c = 6$, $b = 7$: $\overline{abc} = 167$
    - For $c = 7$, $b = 8$: $\overline{abc} = 178$
    - For $c = 8$, $b = 9$: $\overline{abc} = 189$

The final answer is $\boxed{112, 123, 134, 145, 156, 167, 178, 189}$