If $$a+b+c=3, a^2+b^2+c^2=6$$​ and $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$$​, where a, b and c $$\not=$$​ 0, then abc = ?

Correct option is D

Given:

a + b + c = 3

​$$a^2 + b^2 + c^2 = 6$$​​

​$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$$​​

We need to find the value of abc, where $$a, b, c \neq 0.$$​​

Formula Used:

$$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$$​

Solution:

Substitute the given values a + b + c = 3 and $$a^2 + b^2 + c^2 = 6:$$​​

​$$6 = (3)^2 - 2(ab + bc + ca)$$​​

6 = 9 - 2(ab + bc + ca)

2(ab + bc + ca) = 3

ab + bc + ca $$= \frac{3}{2}$$​

We are also given:

​$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$$​

​$$\frac{ bc + ca+ab}{abc} = 1$$​

Substitute ab + bc + ca = $$\frac{3}{2}$$​ into this equation:

​$$\frac{\frac{3}{2}}{abc} = 1$$​

​$$abc = \frac{3}{2}$$​