CUET UG 2025 — Mathematics Algebra

Consider the sum of elements in the first row:

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

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

$$= 0$$=0

Similarly, the sum of elements in the second row:

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

And the sum of elements in the third row:

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

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Since each row sums to zero, applying the column operation $C_1 + C_2 + C_3 \to C_3$C1​+C2​+C3​→C3​:

$$\begin{vmatrix} a-b & b-c & 0 \\ b-c & c-a & 0 \\ c-a & a-b & 0 \end{vmatrix}$$​a−bb−cc−a​b−cc−aa−b​000​​

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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:

$$\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} = 0$$​a−bb−cc−a​b−cc−aa−b​c−aa−bb−c​​=0

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The condition that $a$a, $b$b, and $c$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$0.