If \(a, b\) and \(c\) are distinct prime numbers then the value of \(\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
In every row the three entries add up to zero:
\((a - b) + (b - c) + (c - a) = 0\)
So the operation \(C_1 \to C_1 + C_2 + C_3\), which does not change the determinant, makes every entry of the first column 0:
\( \displaystyle \Delta = \begin{vmatrix}0&b-c&c-a\\0&c-a&a-b\\0&a-b&b-c\end{vmatrix}\)
A determinant with a column of zeros is 0, so \(\Delta = 0\). This holds for any a, b and c; the fact that they are distinct primes plays no part.