$$24=\sum_{cyc}(a-2)^2=\sum_{cyc}a^2-4\sum_{cyc}a+4\cdot 6\implies \sum_{cyc}a^2=4\sum_{cyc}a=80$$AM-QM: $$\sqrt{\frac{80-d^2}{5}}=\sqrt{\frac{a^2+b^2+c^2+e^2+f^2}{5}}\ge \frac{|a|+|b|+|c|+|e|+|f|}{5}\ge\left|\frac{a+b+c+e+f}{5}\right|=\left|\frac{20-d}{5}\right|$$Squaring both sides gives $$d(20-3d)\ge 0$$$$\max d\le\frac{20}{3}$$Numbers $$a=b=c=e=f=\frac{8}{3},\ d=\frac{20}{3}$$satisfy both given equations thus $$\max d\ge\frac{20}{3}$$Hence $$\max d=\frac{20}{3}$$.