If `a,b,c in R` then number of ordered triplet `(a,b,c)` which satisfy the equations `a^(2)+2b=6,b^(2)+4c=-7` and `c^(2)+6a=-13` is

Options

A. 0
B. 1
C. 2
D. More than 2

Explanation

To solve the problem, we need to find the number of ordered triplets \((a, b, c)\) that satisfy the following equations:

1. \(a^2 + 2b = 6\) (Equation 1)
2. \(b^2 + 4c = -7\) (Equation 2)
3. \(c^2 + 6a = -13\) (Equation 3)

Let's go through the solution step by step.

### Step 1: Rearranging the equations
We can rewrite each equation to express \(b\), \(c\), and \(a\) in terms of the others.

From Equation 1:
\[ 2b = 6 - a^2 \implies b = \frac{6 - a^2}{2} \]

From Equation 2:
\[ 4c = -7 - b^2 \implies c = \frac{-7 - b^2}{4} \]

From Equation 3:
\[ 6a = -13 - c^2 \implies a = \frac{-13 - c^2}{6} \]

### Step 2: Substituting values
Now we will substitute the expression for \(b\) from Equation 1 into Equation 2.

Substituting \(b\) into Equation 2:
\[ b^2 = \left(\frac{6 - a^2}{2}\right)^2 = \frac{(6 - a^2)^2}{4} \]
So,
\[ 4c = -7 - \frac{(6 - a^2)^2}{4} \implies c = \frac{-7 - \frac{(6 - a^2)^2}{4}}{4} \]

### Step 3: Substitute \(c\) into Equation 3
Now substitute the expression for \(c\) into Equation 3:
\[ c^2 = \left(\frac{-7 - \frac{(6 - a^2)^2}{4}}{4}\right)^2 \]
Substituting this into Equation 3 gives us a complex equation in terms of \(a\).

### Step 4: Finding a common expression
Instead of solving the complex equation directly, we can add all three equations:
\[ (a^2 + 2b) + (b^2 + 4c) + (c^2 + 6a) = 6 - 7 - 13 \]
This simplifies to:
\[ a^2 + b^2 + c^2 + 2b + 4c + 6a = -14 \]

### Step 5: Completing the square
We can complete the square for each variable:
\[ (a^2 + 6a) + (b^2 + 2b) + (c^2 + 4c) = -14 \]
Completing the square:
\[ (a + 3)^2 - 9 + (b + 1)^2 - 1 + (c + 2)^2 - 4 = -14 \]
This leads to:
\[ (a + 3)^2 + (b + 1)^2 + (c + 2)^2 = 0 \]

### Step 6: Analyzing the equation
Since the sum of squares is equal to zero, each square must individually be zero:
\[ (a + 3)^2 = 0 \implies a + 3 = 0 \implies a = -3 \]
\[ (b + 1)^2 = 0 \implies b + 1 = 0 \implies b = -1 \]
\[ (c + 2)^2 = 0 \implies c + 2 = 0 \implies c = -2 \]

### Conclusion
The only solution we found is \((a, b, c) = (-3, -1, -2)\).

### Final Answer
Thus, the number of ordered triplets \((a, b, c)\) that satisfy the equations is **1**.

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