Explanation of the steps:

Call `E` $\in$ `CD` easy to prove `60^0 `$ = ((\widehat{(SCD);(ABCD)}) = \widehat{SEI}$

From that 

`SI = tan 60^0.EI = a\sqrt{3}` So `V_{S.ABCD} = 1/3a\sqrt{3}.a^2 = {a^3\sqrt{3}}/3`

`BC =a;`$\sqrt{({2a}/3)^2 + (a\sqrt{3})^2}$` = {a\sqrt{31}}/3;SC=`$\sqrt{SI^2 + CB^2 + BI^2}$ `= {2\sqrt{10}a}/3`

We see `AD//BC` so `d(AD;SC)=d(AD;(SBC))=d(D;(SBC))` we observe the pyramid `S.BCD` whose volume is `V_{S.BCD} = 1/3.a\sqrt{3}.{a^2}/2 = {a^3\sqrt{3}}/6`so to find the distance`d(D;(SBC))` we need to find the area $\triangle$`SBC`

We have :

`BC = a;SB = `$\sqrt{({2a}/3)^2 + (a\sqrt{3})^2}$`={a\sqrt{31}}/3;SC=`$\sqrt{SI^2+CB^2+BI^2 }$`={2\sqrt{10}a}/3`

Hence the area :

`S_{\triangleSBC} =`$\sqrt{p(p - SB)(p-SC)(p-BC)}$`;(p ={a + {a\sqrt{31}}/3+ {2\sqrt{10}a}/3}/2)={\sqrt{31}}/6.a^2`

So `d(AD;SC) = d(D;(SBC)) = {3.V_{S.BCD}}/{S_{\triangleBCD}`` ={3\sqrt{93}}/{31}a`