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(A) Let the given determinant be $\Delta = 0$. We can split the determinant using the property of linearity in the third row:$\Delta = \left|\begin{ar

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$a, b$ and $c$ must be equal

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(A) Let the given determinant be $\Delta = 0$. We can split the determinant using the property of linearity in the third row:$\Delta = \left|\begin{array}{lll}{a-b} & {b-c} & {c-a} \ {b-c} & {c-a} & {a-b} \ {c-a} & {a-b} & {b-c}\end{array}ight| + \left|\begin{array}{ccc}{0} & {b-c} & {c-a} \ {0} & {c-a} & {a-b} \ {1} & {a-b} & {b-c}\end{array}ight| = 0$The first determinant is $0$ because the sum of its columns is $0$ (or by observing it is a cyclic determinant which vanishes). For the second determinant,expanding along the first column:$1 \cdot ((b-c)(a-b) - (c-a)(c-a)) = 0$$(b-c)(a-b) - (c-a)^2 = 0$$ab - b^2 - ac + bc - (c^2 + a^2 - 2ac) = 0$$ab - b^2 - ac + bc - c^2 - a^2 + 2ac = 0$$ab + bc + ac - a^2 - b^2 - c^2 = 0$$a^2 + b^2 + c^2 - ab - bc - ca = 0$Multiplying by $2$:$(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$Since $a, b, c \in R$,this implies $a-b=0, b-c=0, c-a=0$,which means $a=b=c$.

If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$,where $a, b, c \in R - \{0\}$,then:

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