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(D) Let $\Delta = \left|\begin{array}{ccc}1+x & 1 & 1 \ 1+y & 1+2 y & 1 \ 1+z & 1+z & 1+3 z\end{array}ight|$.Applying $C_1 	o C_1 - C_3$ and $C_2

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(D) Let $\Delta = \left|\begin{array}{ccc}1+x & 1 & 1 \ 1+y & 1+2 y & 1 \ 1+z & 1+z & 1+3 z\end{array}ight|$.Applying $C_1 	o C_1 - C_3$ and $C_2 	o C_2 - C_3$:$\Delta = \left|\begin{array}{ccc}x & 0 & 1 \ y & 2y & 1 \ z-3z & -2z & 1+3z\end{array}ight| = \left|\begin{array}{ccc}x & 0 & 1 \ y & 2y & 1 \ -2z & -2z & 1+3z\end{array}ight|$.Taking $x, y, z$ common from $C_1, C_2, C_3$ respectively:$\Delta = xyz \left|\begin{array}{ccc}1+1/x & 1/x & 1/x \ 1/y+1 & 1/y+2 & 1/y \ 1/z+1 & 1/z+1 & 1/z+3\end{array}ight|$.Alternatively,expand the determinant directly:$\Delta = (1+x)[(1+2y)(1+3z) - (1+z)] - 1[(1+y)(1+3z) - (1+z)] + 1[(1+y)(1+z) - (1+2y)(1+z)]$.After simplification,$\Delta = x(2y)(3z) + x(2y) + x(3z) + (2y)(3z) = 6xyz + 2xy + 3xz + 6yz$.Factor out $xyz$: $\Delta = xyz(6 + \frac{2}{z} + \frac{3}{y} + \frac{6}{x})$.Comparing with $10kxyz(3 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})$,we find $k=1$.

If $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right| = 10 k x y z \left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,then $k = \text{ . . . . . . }$ (where $x, y, z \neq 0$ and $3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \neq 0$).

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