The determinant left| begin{array}{ccc} ^x{C_ | vedclass

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(C) We know that $^n{C_r} = \frac{n(n-1)...(n-r+1)}{r!}$.Substituting these values into the determinant:$\Delta = \left| \begin{array}{ccc} x & \frac{

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$\frac{1}{12} xyz (x - y) (y - z) (z - x)$

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(C) We know that $^n{C_r} = \frac{n(n-1)...(n-r+1)}{r!}$.Substituting these values into the determinant:$\Delta = \left| \begin{array}{ccc} x & \frac{x(x-1)}{2} & \frac{x(x-1)(x-2)}{6} \ y & \frac{y(y-1)}{2} & \frac{y(y-1)(y-2)}{6} \ z & \frac{z(z-1)}{2} & \frac{z(z-1)(z-2)}{6} \end{array} ight|$Taking $\frac{x}{1}, \frac{y}{2}, \frac{z}{6}$ common from the columns respectively:$\Delta = \frac{xyz}{12} \left| \begin{array}{ccc} 1 & x-1 & (x-1)(x-2) \ 1 & y-1 & (y-1)(y-2) \ 1 & z-1 & (z-1)(z-2) \end{array} ight|$Applying $C_2 ightarrow C_2 + C_1$:$\Delta = \frac{xyz}{12} \left| \begin{array}{ccc} 1 & x & x^2-3x+2 \ 1 & y & y^2-3y+2 \ 1 & z & z^2-3z+2 \end{array} ight|$Using column operations $C_3 ightarrow C_3 + 3C_2 - 2C_1$,we reduce the determinant to the standard Vandermonde form:$\Delta = \frac{xyz}{12} \left| \begin{array}{ccc} 1 & x & x^2 \ 1 & y & y^2 \ 1 & z & z^2 \end{array} ight| = \frac{xyz}{12} (x-y)(y-z)(z-x)$.

The determinant $\left| \begin{array}{ccc} ^x{C_1} & ^x{C_2} & ^x{C_3} \\ ^y{C_1} & ^y{C_2} & ^y{C_3} \\ ^z{C_1} & ^z{C_2} & ^z{C_3} \end{array} \right|$ equals:

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