$\text{Value of}\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|=$

The correct option is 0.

To find the value of the given determinant, we can use the properties of determinants to simplify the matrix. The given determinant is:

$$\left|\begin{array}{ccc}a-b& b-c& c-a\\ b-c& c-a& a-b\\ c-a& a-b& b-c\end{array}\right|$$

Let's apply the column operation where we add all columns to the first column: C₁ → C₁ + C₂ + C₃.
This operation does not change the value of the determinant.

Let's calculate the new elements of the first column:

For the first row:
$(a-b)+(b-c)+(c-a)=a-b+b-c+c-a=0$

For the second row:
$(b-c)+(c-a)+(a-b)=b-c+c-a+a-b=0$

For the third row:
$(c-a)+(a-b)+(b-c)=c-a+a-b+b-c=0$

Applying this column transformation, the determinant becomes:

$$\left|\begin{array}{ccc}0& b-c& c-a\\ 0& c-a& a-b\\ 0& a-b& b-c\end{array}\right|$$

A standard property of determinants states that if all elements of any single row or column of a determinant are zero, then the value of the determinant is equal to zero.
Since every element in the first column is zero, the value of the entire determinant is indeed 0.