Let $t$ be a positive constant. Given two points $A(2t,\ 2t,\ 0),\ B(0,\ 0,\ t)$ in a space with the origin $O$. Suppose mobile points $P$ in such way that $\overrightarrow{OP}\cdot \overrightarrow{AP}+\overrightarrow{OP}\cdot \overrightarrow{BP}+\overrightarrow{AP}\cdot \overrightarrow{BP}=3.$
Find the value of $t$ such that the maximum value of $OP$ is 3.
1. **Given Points and Condition:**
   - Points \( A(2t, 2t, 0) \) and \( B(0, 0, t) \).
   - Condition: \(\overrightarrow{OP} \cdot \overrightarrow{AP} + \overrightarrow{OP} \cdot \overrightarrow{BP} + \overrightarrow{AP} \cdot \overrightarrow{BP} = 3\).

2. **Vector Definitions:**
   - Let \( P(x, y, z) \).
   - Vectors:
     \[
     \overrightarrow{OP} = (x, y, z)
     \]
     \[
     \overrightarrow{AP} = (x - 2t, y - 2t, z)
     \]
     \[
     \overrightarrow{BP} = (x, y, z - t)
     \]

3. **Dot Products:**
   - Calculate the dot products:
     \[
     \overrightarrow{OP} \cdot \overrightarrow{AP} = x(x - 2t) + y(y - 2t) + z \cdot z = x^2 - 2tx + y^2 - 2ty + z^2
     \]
     \[
     \overrightarrow{OP} \cdot \overrightarrow{BP} = x^2 + y^2 + z(z - t) = x^2 + y^2 + z^2 - tz
     \]
     \[
     \overrightarrow{AP} \cdot \overrightarrow{BP} = (x - 2t)x + (y - 2t)y + z(z - t) = x^2 - 2tx + y^2 - 2ty + z^2 - tz
     \]

4. **Sum of Dot Products:**
   - Sum the dot products:
     \[
     \overrightarrow{OP} \cdot \overrightarrow{AP} + \overrightarrow{OP} \cdot \overrightarrow{BP} + \overrightarrow{AP} \cdot \overrightarrow{BP} = (x^2 - 2tx + y^2 - 2ty + z^2) + (x^2 + y^2 + z^2 - tz) + (x^2 - 2tx + y^2 - 2ty + z^2 - tz)
     \]
     \[
     = 3x^2 + 3y^2 + 3z^2 - 4tx - 4ty - 2tz
     \]
     Given that this sum equals 3:
     \[
     3(x^2 + y^2 + z^2) - 4t(x + y) - 2tz = 3
     \]

5. **Simplify and Solve for \( t \):**
   - Let \( A = x^2 + y^2 + z^2 \):
     \[
     3A - 4t(x + y) - 2tz = 3
     \]
   - Assume \( OP = 3 \), then \( A = 9 \):
     \[
     3(9) - 4t(x + y) - 2tz = 3
     \]
     \[
     27 - 4t(x + y) - 2tz = 3
     \]
     \[
     24 = 4t(x + y) + 2tz
     \]
     \[
     12 = 2t(x + y + z)
     \]
     \[
     6 = t(x + y + z)
     \]
     \[
     t = \frac{6}{x + y + z}
     \]

6. **Maximize \( OP \):**
   - Given \( OP = 3 \), we need to find \( t \) such that the maximum value of \( OP \) is 3.
   - From the previous steps, we have:
     \[
     t = \frac{6}{x + y + z}
     \]
   - To maximize \( OP \), we need \( x + y + z = 6 \):
     \[
     t = \frac{6}{6} = 1
     \]

The final answer is \( \boxed{ t = 1 } \)