Combined 100 year birthday
Mathematica 138 107 237 bytes
My first attempts were sloppy and unworthy. This should work generally, provided that the birth dates are given in order of earliest to latest, as in the test cases.
The test cases were well-chosen and difficult to implement correctly.
The code turned out way longer than I had hoped.
The general logic is to

Solve for x (in days) such that the distance from x to each date of birth sums to the target passage of time (e.g 100 years converted to days).
If the earliest date + x days &gt; latest date of birth, remove the latest date of birth and repeat step (1), Else move to (3).
Return (earliest date + x days)


t_~f~d_:=If[DateDifference[a=DateObject[DatePlus[d[[1]],Solve[t 365.25==Tr@(x-Join[{0},
QuantityMagnitude@(DateDifference[d[[1]],#,&quot;Day&quot;]&amp;/@Rest@d)]),{x}][[1,1,2]]]/.
{y_,m_,n_,h_}:&gt;{y,m,n+1}],d[[-1]]]&gt;Quantity[0,&quot;Days&quot;],f[t,Most@d],a]


Test Cases
f[10, {{2000, 1, 1}}]
f[100, {{1975, 3, 5}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]
f[100, {{1975, 3, 6}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]
f[100, {{1975, 3, 9}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]
f[100, {{1975, 3, 10}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12,  20}}]
f[3, {{1990, 8, 1}, {1995, 7, 2}}]
f[50, {{1975, 3, 5}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]