If a, b, c are distinct and rational numbers, | vedclass

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(B) Let $S = ab + bc + ca$ and $K = a^2 + b^2 + c^2$. The determinant is $\Delta = \left| \begin{array}{ccc} K & S & S \ S & K & S \ S & S & K \end{

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Rational and Positive

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(B) Let $S = ab + bc + ca$ and $K = a^2 + b^2 + c^2$. The determinant is $\Delta = \left| \begin{array}{ccc} K & S & S \ S & K & S \ S & S & K \end{array} ight|$.Applying $R_1 	o R_1 + R_2 + R_3$,we get $\Delta = (K + 2S) \left| \begin{array}{ccc} 1 & 1 & 1 \ S & K & S \ S & S & K \end{array} ight|$.Since $K + 2S = (a+b+c)^2$,we have $\Delta = (a+b+c)^2 \left| \begin{array}{ccc} 1 & 1 & 1 \ S & K & S \ S & S & K \end{array} ight|$.Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$,we get $\Delta = (a+b+c)^2 \left| \begin{array}{ccc} 1 & 0 & 0 \ S & K-S & 0 \ S & 0 & K-S \end{array} ight| = (a+b+c)^2 (K-S)^2$.Substituting back,$K-S = a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2} [(a-b)^2 + (b-c)^2 + (c-a)^2]$.Thus,$\Delta = (a+b+c)^2 	imes \left( \frac{1}{2} [(a-b)^2 + (b-c)^2 + (c-a)^2] ight)^2$.Since $a, b, c$ are distinct rational numbers,$(a-b)^2, (b-c)^2, (c-a)^2$ are positive rational numbers. Therefore,$\Delta$ is a positive rational number.

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