d
(d) \({{\log x} \over {b - c}} = {{\log y} \over {c - a}} = {{\log z} \over {a - b}} = k\,({\rm{say}})\)

\( \Rightarrow \)\(\log x = k(b - c),\,\log y = k(c - a),\,\log z = k(a - b)\)

\( \Rightarrow \)\(x = {e^{k(b - c)}},\,y = {e^{k(c - a)}},\,z = {e^{k(a - b)}}\)

\(\therefore xyz = {e^{k(b - c) + k(c - a) + k(a - b)}} = {e^0} = 1\)

\({x^a}{y^b}{z^c} = {e^{k(b - c)a + k(c - a)b + k(a - b)c}} = {e^0} = 1 = xyz\)

\({x^{b + c}}{y^{c + a}}{z^{a + b}} = {e^{k({b^2} - {c^2}) + k({c^2} - {a^2}) + k({a^2} - {b^2})}} = {e^0} = 1\).