If a+b+c=3,`a^2+b^2+c^2=6` and `1/a+1/b+1/c=1` where a,b,c are all non zero then abc is:

Options

A. `(1/3)`
B. `(2/3)`
C. `(3/2)`
D. `1`

Answer

To solve the problem step by step, we will use the given equations systematically.

**Step 1: Write down the given equations.**
We have:
1. \( a + b + c = 3 \) (Equation 1)
2. \( a^2 + b^2 + c^2 = 6 \) (Equation 2)
3. \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \) (Equation 3)

**Step 2: Use the identity for the square of a sum.**
From Equation 1, we can use the identity:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting the values from Equations 1 and 2:
\[ 3^2 = 6 + 2(ab + ac + bc) \]
This simplifies to:
\[ 9 = 6 + 2(ab + ac + bc) \]

**Step 3: Simplify to find \( ab + ac + bc \).**
Rearranging the equation gives:
\[ 9 - 6 = 2(ab + ac + bc) \]
\[ 3 = 2(ab + ac + bc) \]
Dividing both sides by 2:
\[ ab + ac + bc = \frac{3}{2} \quad (Equation 4) \]

**Step 4: Use Equation 3 to express \( ab + ac + bc \) in terms of \( abc \).**
From Equation 3, we know:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} = 1 \]
This implies:
\[ ab + ac + bc = abc \]
Substituting from Equation 4:
\[ abc = \frac{3}{2} \quad (Equation 5) \]

**Step 5: Conclusion.**
Thus, the value of \( abc \) is:
\[ abc = \frac{3}{2} \]

**Final Answer:**
The value of \( abc \) is \( \frac{3}{2} \).

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