(d) $\left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z + x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}\,} \right|$ $ = \left| {\,\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{2y}&{2x}&0\\{2z}&0&{2x}\end{array}\,} \right|$
${R_2} \to {R_2} + {R_1}$ and ${R_3} \to {R_3} + {R_1}$
$ = 4\left| {\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\y&x&0\\z&0&x\end{array}} \right|$
$ = 4[(y + z)({x^2}) - (x - z)(xy)$ $ + (x - y)( - zx)]$
$ = 4[{x^2}y + z{x^2} - {x^2}y + xyz - z{x^2} + xyz]$ $ = 8xyz$
Hence, $k = 8$.