Let ` (log a)/( b-c) = (logb)/(c-a) = (log c)/(a-b) `
STATEMENT 1: ` a^(a) b^(b) c^(c) = 1`
STATEMENT 2: `a^(b+c) b^(c +a) c^(a + b) = 1`

Options

A. Statemant-1 is True , Statement-2 is True, Statement -2 is a correct explanation for Statement-1
B. Statemant-1 is True , Statement-2 is True, Statement -2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Stetement-2 is False.
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem step by step, we start with the given equation:

\[ \frac{\log a}{b - c} = \frac{\log b}{c - a} = \frac{\log c}{a - b} \]

Let’s denote this common value as \( k \). Thus, we can write:

\[ \log a = k(b - c) \]
\[ \log b = k(c - a) \]
\[ \log c = k(a - b) \]

### Step 1: Add the three equations
Adding the three equations gives us:

\[ \log a + \log b + \log c = k(b - c) + k(c - a) + k(a - b) \]

### Step 2: Simplify the right-hand side
The right-hand side simplifies as follows:

\[ k(b - c + c - a + a - b) = k(0) = 0 \]

### Step 3: Conclude the left-hand side
Thus, we have:

\[ \log a + \log b + \log c = 0 \]

This implies:

\[ \log(abc) = 0 \]

### Step 4: Exponentiate to remove logarithm
Exponentiating both sides gives:

\[ abc = 1 \]

### Step 5: Verify Statement 1
Now we need to verify Statement 1:

\[ a^a b^b c^c = 1 \]

Using the property of logarithms, we can express this as:

\[ \log(a^a b^b c^c) = a \log a + b \log b + c \log c \]

Substituting the values of \(\log a\), \(\log b\), and \(\log c\):

\[ = a(k(b - c)) + b(k(c - a)) + c(k(a - b)) \]

### Step 6: Simplify the expression
This simplifies to:

\[ = k(ab - ac + bc - ab + ca - cb) = k(0) = 0 \]

Thus:

\[ \log(a^a b^b c^c) = 0 \implies a^a b^b c^c = 1 \]

So, Statement 1 is true.

### Step 7: Verify Statement 2
Now we verify Statement 2:

\[ a^{b+c} b^{c+a} c^{a+b} = 1 \]

Using the property of logarithms again:

\[ \log(a^{b+c} b^{c+a} c^{a+b}) = (b+c) \log a + (c+a) \log b + (a+b) \log c \]

### Step 8: Substitute the values
Substituting the values of \(\log a\), \(\log b\), and \(\log c\):

\[ = (b+c)(k(b-c)) + (c+a)(k(c-a)) + (a+b)(k(a-b)) \]

### Step 9: Simplify the expression
This simplifies to:

\[ = k[(b+c)(b-c) + (c+a)(c-a) + (a+b)(a-b)] \]

### Step 10: Expand and combine terms
Expanding each term and combining gives:

\[ = k[0] = 0 \]

Thus:

\[ \log(a^{b+c} b^{c+a} c^{a+b}) = 0 \implies a^{b+c} b^{c+a} c^{a+b} = 1 \]

So, Statement 2 is also true.

### Conclusion
Both statements are true, and we can conclude:

- Statement 1: \( a^a b^b c^c = 1 \) is true.
- Statement 2: \( a^{b+c} b^{c+a} c^{a+b} = 1 \) is true.