If the line `x=y=z` intersect the lines `xsinA+ysinB+zsinC-2d^(2)=0=xsin(2A)+ysin(2B)+zsin(2C)-d^(2),` where `A, B, C` are the internal angles of a triangle and `"sin"(A)/(2)"sin"(B)/(2)"sin"(C)/(2)=k` then the value of `64k` is equal to

Answer

To solve the given problem, we will follow these steps:

### Step 1: Set up the equations
We have two equations based on the intersection of the line \( x = y = z \) with the given lines:
1. \( x \sin A + y \sin B + z \sin C - 2d^2 = 0 \)
2. \( x \sin(2A) + y \sin(2B) + z \sin(2C) - d^2 = 0 \)

Since \( x = y = z \), we can denote \( x = y = z = \lambda \). Thus, we can rewrite the equations as:
1. \( \lambda \sin A + \lambda \sin B + \lambda \sin C = 2d^2 \)
2. \( \lambda \sin(2A) + \lambda \sin(2B) + \lambda \sin(2C) = d^2 \)

### Step 2: Simplify the equations
Factoring out \( \lambda \) from both equations gives us:
1. \( \lambda (\sin A + \sin B + \sin C) = 2d^2 \)
2. \( \lambda (\sin(2A) + \sin(2B) + \sin(2C)) = d^2 \)

### Step 3: Divide the equations
Now, we can divide the first equation by the second equation:
\[ \frac{\sin(2A) + \sin(2B) + \sin(2C)}{\sin A + \sin B + \sin C} = \frac{d^2}{2d^2} = \frac{1}{2} \]

### Step 4: Use the sine double angle formula
Using the double angle identity, we know that:
\[ \sin(2A) = 2 \sin A \cos A \]
Thus, we can rewrite:
\[ \sin(2A) + \sin(2B) + \sin(2C) = 2(\sin A \cos A + \sin B \cos B + \sin C \cos C) \]

### Step 5: Substitute and simplify
Substituting back into our equation gives:
\[ \frac{2(\sin A \cos A + \sin B \cos B + \sin C \cos C)}{\sin A + \sin B + \sin C} = \frac{1}{2} \]
This simplifies to:
\[ \sin A \cos A + \sin B \cos B + \sin C \cos C = \frac{1}{4}(\sin A + \sin B + \sin C) \]

### Step 6: Use the identity for angles in a triangle
Using the identity for angles in a triangle:
\[ \sin A + \sin B + \sin C = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]
we can substitute this into our equation.

### Step 7: Solve for k
From our earlier steps, we can derive:
\[ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16} \]
Thus, we have:
\[ k = \frac{1}{16} \]

### Step 8: Calculate \( 64k \)
Finally, we need to find \( 64k \):
\[ 64k = 64 \times \frac{1}{16} = 4 \]

### Final Answer
The value of \( 64k \) is \( 4 \).

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