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 equal to

Options

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

Answer

To solve for \( abc \) given the equations:

1. \( a + b + c = 3 \)
2. \( a^2 + b^2 + c^2 = 6 \)
3. \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1 \)

we will follow these steps:

### Step 1: Use the identity for the square of a sum

We know that:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]
Substituting the known values:
\[ 3^2 = 6 + 2(ab + bc + ca) \]
This simplifies to:
\[ 9 = 6 + 2(ab + bc + ca) \]
Subtracting 6 from both sides gives:
\[ 3 = 2(ab + bc + ca) \]
Dividing by 2:
\[ ab + bc + ca = \frac{3}{2} \]

### Step 2: Use the third equation

From the third equation, we can express it in terms of \( abc \):
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + bc + ca}{abc} = 1 \]
Substituting \( ab + bc + ca \):
\[ \frac{\frac{3}{2}}{abc} = 1 \]
This implies:
\[ abc = \frac{3}{2} \]

### Step 3: Conclusion

Thus, the value of \( abc \) is:
\[ \boxed{\frac{3}{2}} \]