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(D) Let $\Delta = \left| \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \ a & \frac{b^2+c^2}{a} & a \ b & b & \frac{c^2+a^2}{b} \end{array} ight|$.T

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$4$

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(D) Let $\Delta = \left| \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \ a & \frac{b^2+c^2}{a} & a \ b & b & \frac{c^2+a^2}{b} \end{array} ight|$.Taking $1/c$ from $R_1$,$1/a$ from $R_2$,and $1/b$ from $R_3$,we get:$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} a^2+b^2 & c^2 & c^2 \ a^2 & b^2+c^2 & a^2 \ b^2 & b^2 & c^2+a^2 \end{array} ight|$.Applying $R_1 ightarrow R_1 - (R_2 + R_3)$:$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} 0 & -2b^2 & -2a^2 \ a^2 & b^2+c^2 & a^2 \ b^2 & b^2 & c^2+a^2 \end{array} ight|$.Applying $R_2 ightarrow R_2 + \frac{1}{2}R_1$ and $R_3 ightarrow R_3 + \frac{1}{2}R_1$:$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} 0 & -2b^2 & -2a^2 \ a^2 & c^2 & 0 \ b^2 & 0 & c^2 \end{array} ight|$.Expanding along $R_1$:$\Delta = \frac{1}{abc} [0 - (-2b^2)(a^2c^2 - 0) + (-2a^2)(0 - b^2c^2)]$$\Delta = \frac{1}{abc} [2a^2b^2c^2 + 2a^2b^2c^2] = \frac{4a^2b^2c^2}{abc} = 4abc$.Given $\Delta = \alpha abc$,we have $\alpha = 4$.

For non-zero,real $a, b$ and $c$,if $\left| \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{array} \right| = \alpha abc$,then the value of $\alpha$ is

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