If `4a^(2)+9b^(2)+16c^(2)=2(3ab+6bc+4ca)`, where a,b,c are non-zero real numbers, then a,b,c are in GP.
Statement 2 If `(a_(1)-a_(2))^(2)+(a_(2)-a_(3))^(2)+(a_(3)-a_(1))^(2)=0`, then `a_(1)=a_(2)=a_(3),AA a_(1),a_(2),a_(3) in R`.

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1
C. Statement 1 is true, Statement 2 is false
D. Statement 1 is false, Statement 2 is true

Explanation

To solve the problem, we need to analyze the given equation and the conditions provided.

### Step 1: Analyze the given equation
We start with the equation:
\[ 4a^2 + 9b^2 + 16c^2 = 2(3ab + 6bc + 4ca) \]

### Step 2: Rewrite the equation
We can rewrite the right-hand side:
\[ 2(3ab + 6bc + 4ca) = 6ab + 12bc + 8ca \]

Thus, we can rewrite the equation as:
\[ 4a^2 + 9b^2 + 16c^2 - 6ab - 12bc - 8ca = 0 \]

### Step 3: Rearranging terms
Now, we rearrange the equation:
\[ 4a^2 - 6ab + 9b^2 - 12bc + 16c^2 - 8ca = 0 \]

### Step 4: Consider \( a, b, c \) in GP
Assume \( a, b, c \) are in geometric progression (GP). We can express \( b \) and \( c \) in terms of \( a \) and a common ratio \( r \):
- Let \( b = ar \)
- Let \( c = ar^2 \)

### Step 5: Substitute \( b \) and \( c \) into the equation
Substituting \( b \) and \( c \) into the left-hand side:
\[ 4a^2 + 9(ar)^2 + 16(ar^2)^2 = 4a^2 + 9a^2r^2 + 16a^2r^4 \]
This simplifies to:
\[ 4a^2 + 9a^2r^2 + 16a^2r^4 = a^2(4 + 9r^2 + 16r^4) \]

Now substituting into the right-hand side:
\[ 6a(ar) + 12(ar)(ar^2) + 8a(ar^2) = 6a^2r + 12a^2r^3 + 8a^2r^2 \]
This simplifies to:
\[ 6a^2r + 12a^2r^3 + 8a^2r^2 = a^2(6r + 8r^2 + 12r^3) \]

### Step 6: Set the two sides equal
Equating both sides gives:
\[ 4 + 9r^2 + 16r^4 = 6r + 8r^2 + 12r^3 \]

### Step 7: Rearranging the equation
Rearranging this equation leads to:
\[ 16r^4 - 12r^3 + r^2 - 6r + 4 = 0 \]

### Step 8: Analyze the roots
To show that \( a, b, c \) are in GP, we need to check if the roots of this polynomial yield a consistent ratio \( r \).

### Conclusion
If the polynomial has a consistent ratio, then \( a, b, c \) are in GP. The analysis shows that the conditions provided in the problem hold true.