If a, b, c are distinct positive numbers, each different from 1, such that \[log_{b}\, a\, log_{c}\, a\, -log_{a}\, a\]+\[log_{a}\, b\, log_{c}\, b\, -log_{b}\, b\]+\[log_{a}\, c\, log_{b}\, c-log_{c}\, c\]= 0 , then \[abc = \]
a) 1
b) 2
c) 3
d) 4

Hint:To solve this question, apply the logarithmic properties such as the \[{\log _n}m = \dfrac{{{{\log }_m}}}{{{{\log }_n}}}\] and
\[{\log _a}a = 1\]. Also, apply the identity \[a^3+b^3+c^3=3abc\] when \[a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c{\text{ }} = {\text{ }}0\] to get the value of \[abc\].

Formula Used:
\[lo{g_a}{\text{ }}mn{\text{ }} = {\text{ }}lo{g_a}{\text{ }}m{\text{ }} + {\text{ }}lo{g_a}\,n\]
 \[lo{g_a}\,m/n{\text{ }} = {\text{ }}lo{g_a}{\text{ }}m{\text{ }} - {\text{ }}lo{g_a}{\text{ }}n\]
\[ log_{a}m^{n}=nlog_{a}m \]

Complete step by step Solution:
Before solving the question, It is important to know a few logarithmic properties which are required further to simplify the equations. A few important logarithmic properties are mentioned below
Product property is given by: \[lo{g_a}{\text{ }}mn{\text{ }} = {\text{ }}lo{g_a}{\text{ }}m{\text{ }} + {\text{ }}lo{g_a}\,n\]
Quotient property is given by: \[lo{g_a}\,m/n{\text{ }} = {\text{ }}lo{g_a}{\text{ }}m{\text{ }} - {\text{ }}lo{g_a}{\text{ }}n\]
Power property is given by: \[log_{a}m^{n}=nlog_{a}m\]
The product property of logarithm leads to addition, the quotient property of logarithm leads to subtraction, and the power property of logarithm leads to multiplication.

Given:
\[log_{b}\, a\, log_{c}\, a\, -log_{a}\, a\]+\[log_{a}\, b\, log_{c}\, b\, -log_{b}\, b\]+\[log_{a}\, c\, log_{b}\, c-log_{c}\, c\] = 0 - eq (1)
To get the value of \[abc\], use the properties of the logarithm
\[{\log _n}m = \dfrac{{{{\log }_m}}}{{{{\log }_n}}}\] and \[{\log _a}a = 1\]
To simplify the given equation, apply the above-mentioned logarithmic properties to eq (1)
\[\left[ {\dfrac{{\log a}}{{\log b}} \times \dfrac{{\log \,a}}{{\log \,c}} - 1} \right] + \left[ {\dfrac{{\log \,b}}{{\log \,a}} \times \dfrac{{\log \,b}}{{\log \,c}} - 1} \right] + \left[ {\dfrac{{\log \,c}}{{\log \,a}} \times \dfrac{{\log \,c}}{{\log \,b}} - 1} \right] = 0\]

Open the brackets and simplify the above equation to get
\[\dfrac{{{{(\log a)}^2}}}{{\log b \times \log c}} + \dfrac{{{{(log\,b)}^2}}}{{\log a \times \log c}} + \dfrac{{{{(\log \,c)}^2}}}{{\log a \times \log b}} = 3\] -- eq (2)

Multiply the above equation (2) throughout by \[loga{\text{ }}logb{\text{ }}logc\]
\[\dfrac{{\log a \times \log b \times \log c \times {{(\log a)}^2}}}{{\log b \times \log c}} + \dfrac{{\log a \times \log b \times \log c \times {{(log\,b)}^2}}}{{\log a \times \log c}} + \dfrac{{\log a \times \log b \times \log c \times {{(\log \,c)}^2}}}{{\log a \times \log b}} = 3\log a\log b\log c\]
Cancel the like terms and multiply the remaining terms to get
\[{(\log a)^3} + {(\log b)^3} + {(\log c)^3} = 3\log a\log b\log c\]-- eq (3)
We know that \[a^3+b^3+c^3=3abc\] when \[a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c{\text{ }} = {\text{ }}0\]
Here \[a = log{\text{ }}a\], \[b = log{\text{ }}b\] and \[c = log{\text{ }}c{\text{ }}\]
The eq (3) satisfies the condition \[a^3+b^3+c^3=3abc\] when \[a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c{\text{ }} = {\text{ }}0\]. Hence it can be written as
\[\log a + \log b + \log c = 0\]
Apply the logarithmic property \[lo{g_a}{\text{ }}mn{\text{ }} = {\text{ }}lo{g_a}{\text{ }}m{\text{ }} + {\text{ }}lo{g_a}\,n\] to the above equation
\[\log abc = 0\]
Thus \[abc = 1\]

Hence, the correct option is a.

Note: To easily solve the logarithmic questions, apply the logarithmic properties in step by step manner. To reduce or simplify the logarithmic equations, the basic properties of the logarithm such as the product rule, the power rule, or the quotient rule are important. These properties of logarithm help in reducing or simplifying complex logarithmic equations.