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

Consider the sum of elements in the first row:

$(a-b) + (b-c) + (c-a)$

$= a - b + b - c + c - a$

$= 0$

Similarly, the sum of elements in the second row:

$(b-c) + (c-a) + (a-b) = 0$

And the sum of elements in the third row:

$(c-a) + (a-b) + (b-c) = 0$

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Since each row sums to zero, applying the column operation $C_1 + C_2 + C_3 \to C_3$:

$\begin{vmatrix} a-b & b-c & 0 \\ b-c & c-a & 0 \\ c-a & a-b & 0 \end{vmatrix}$

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When an entire column of a determinant consists of zeros, the determinant equals zero.

Since the third column is entirely zero:

$\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} = 0$

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The condition that $a$, $b$, and $c$ are distinct prime numbers does not affect this result. The algebraic property that each row sums to zero holds for any distinct values.

Therefore, the value of the determinant is $0$.