Question 158798
Well, it's both. 
Here's where they are at first.
{{{drawing( 300, 300, -10, 10, -10, 10,
line(0,-6,6,0),
line(6,0,0,6),
circle(0,-6,.2),
circle(0,6,.2),
locate(-.5,-6.5, 1st) ,
locate(-.5,7.2, 3rd) ,
locate(6.2,0.5, 2nd),
locate(3.5,4, 90 ft),
locate(3.5,-3.5, 90 ft)  )}}}
If we set up a coordinate system at first base then the person at first base is at (0,0). 
The x-axis runs from first to second.
The y-axis runs from first to home plate.
The person at third base is at (90,90).
The position of the person at first changes with t by
(10t,0)
and the position of the person on third base changes with t by
(90-6t,90).
The x distance is one leg of the triangle, determined with rate and time.
The y distance, fixed at 90 feet, is another leg of the triangle.
The distance is the hypotenuse of the triangle.
The distance between the two, using the distance formula, is
{{{D^2=(10t-(90-6t))^2+(0-90)^2}}}
{{{D^2=(10t-90+6t))^2+(-90)^2}}}
{{{D^2=(16t-90))^2+8100}}}
{{{D^2=256t^2-2880t+8100+8100}}}
{{{D^2=256t^2-2880t+16200}}}
Find t when D=270.
{{{270^2=256t^2-2880t+16200}}}
{{{72900=256t^2-2880t+16200}}}
{{{256t^2-2880t-56700=0}}}
Use the quadratic formula,
{{{t = (-(-2880) +- sqrt( (-2880)^2-4*256*(-56700) ))/(2*256) }}}
{{{t = (2880 +- sqrt( 8294400+58060800 ))/(512) }}}
Only use the positive value,
{{{t = (2880 + sqrt( 66355200))/(512) }}}
{{{t = (2880 + 8145.9)/(512) }}}
{{{t = 21.5 }}}
Let's check the answer.
After 21.5 seconds, the person on 1st will be at
(10t,0)=(10(21.5),0)=(215,0)
and the person on third will be at
(90-6t,90)=(90-6(21.5),90)=(-39,60)
The distance is then
{{{D^2=(215-(-39))^2+(0-90)^2}}}
{{{D^2=(254)^2+(90)^2}}}
{{{D^2=64516+8100(254)^2+(90)^2}}}
{{{D^2=72616}}}
{{{D=269.5}}}
Good answer.