In the triangle ABC, the opposite sides of angles A,B,C are a,b,c, a square +b square +c square =2 square root 3bcsinA, and point D is in the plane ABC, where DC=2DB, a=6, then what is the maximum value of DA

To solve the problem, we need to find the maximum value of \( DA \) given the conditions. Let's break down the problem step by step.

### Step 1: Given Information
1. **Sides of the triangle:**
   - Opposite to angle \( A \): \( a = 6 \)
   - Opposite to angle \( B \): \( b \)
   - Opposite to angle \( C \): \( c \)
   
2. **Equation relating the sides and angle \( A \):**
   \[
   a^2 + b^2 + c^2 = 2 \sqrt{3} \cdot bc \cdot \sin A
   \]
   
3. **Point \( D \) in the plane:**
   \[
   DC = 2DB
   \]

### Step 2: Simplify and Analyze the Given Equation
Substitute \( a = 6 \) into the given equation:
\[
6^2 + b^2 + c^2 = 2 \sqrt{3} \cdot bc \cdot \sin A
\]
This simplifies to:
\[
36 + b^2 + c^2 = 2 \sqrt{3} \cdot bc \cdot \sin A
\]

### Step 3: Identify Maximum Value of \( DA \)
To find the maximum value of \( DA \), we need to explore the geometric configuration given by \( DC = 2DB \).

Since \( D \) is in the plane of \( \triangle ABC \) and \( DC = 2DB \), let's assume \( D \) lies on the angle bisector of \( \angle ACB \). In this case, \( D \) divides \( BC \) internally in the ratio \( 1:2 \).

### Step 4: Use Trigonometry and Geometry
Using the law of cosines in \( \triangle ABD \) and \( \triangle ACD \), we can express \( DA \) in terms of \( b \), \( c \), and the angle between them. However, without loss of generality, we can simplify the task by maximizing \( DA \) directly from the relation between the points.

### Step 5: Calculating Distances
For simplicity, assume \( \triangle ABC \) is such that \( \angle A = 60^\circ \) since \( \sin A = \sqrt{3}/2 \) simplifies the problem. Given \( a = 6 \), we have:
\[
36 + b^2 + c^2 = 2 \sqrt{3} \cdot bc \cdot \frac{\sqrt{3}}{2} = 3bc
\]
This implies:
\[
b^2 + c^2 + 36 = 3bc
\]
To maximize \( DA \), observe that \( DA \) is maximized when \( D \) lies on the circumcircle of \( \triangle ABC \).

### Step 6: Solving for Maximum \( DA \)
In the context of the triangle with given sides and the relation \( DC = 2DB \):
\[
D \text{ is on the circumcircle of } \triangle ABC
\]
The maximum distance \( DA \) will be the length of the diameter of the circumcircle:
\[
DA \leq 2R \text{ (where \( R \) is the circumradius)}
\]

Using the circumradius formula for a triangle:
\[
R = \frac{abc}{4K}
\]
Given the simplified relations and that \( \angle A = 60^\circ \), we can infer a maximum diameter as a function of the triangle's configuration.

For an equilateral triangle (the simplest case satisfying all conditions with max distances), \( DA \) is maximized when it is \( a\sqrt{3} \).

### Conclusion
So, the maximum value of \( DA \) considering the maximum geometric configuration and circumradius relations:
\[
DA \leq 2R \leq 2 \times \text{(circumradius derived as function of the sides)}
\]
For a triangle where \( a = 6 \) and other values maximized,
\[
DA \leq 6\sqrt{3}
\]

Therefore, the maximum value of \( DA \) is \( \boxed{6\sqrt{3}} \).

Would you like more details on this problem or any specific part? Here are some related questions you might find interesting:

1. How do you derive the circumradius of a triangle given its sides?
2. Can you explain the Law of Cosines in detail?
3. What is the significance of the point \( D \) lying on the angle bisector?
4. How does one maximize distances in a geometric configuration?
5. What are the properties of equilateral triangles in this context?

**Tip:** When solving complex geometric problems, visualizing the configuration and leveraging symmetrical properties can significantly simplify the calculations.

Explore how to maximize the distance DA within triangle ABC, given specific conditions. Learn about the application of the Law of Cosines, properties of triangles, and the circumradius formula in solving this geometric problem. Suitable for advanced high school students and beyond.

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