{QUOTE
{SET Delay 3}
{SET a{=INT((14-{DATE \@ M})/12)}}
{SET b{={DATE \@ yyyy}+4800-a}}
{SET c{={DATE \@ M}+12*a-3}}
{SET d{DATE \@ d}}
{SET jd{=d+INT((153*c+2)/5)+365*b+INT(b/4)-INT(b/100)+INT(b/400)-32045+Delay}}
{SET e{=INT((4*(jd+32044)+3)/146097)}}
{SET f{=jd+32044-INT(146097*e/4)}}
{SET g{=INT((4*f+3)/1461)}}
{SET h{=f-INT(1461*g/4)}}
{SET i{=INT((5*h+2)/153)}}
{SET dd{=h-INT((153*i+2)/5)+1}}
{SET mm{=i+3-12*INT(i/10)}}
{SET yy{=100*e+g-4800+INT(i/10)}}
"{mm}-{dd}-{yy}" \@ "MM/dd/yyyy"}


You cannot simply add 3 to the date or (say) 30th December + 3 would be 32nd December.
You can insert date fields that will give the required values for today's date for the three dates (see attached), but it will not work with typed text dates. You cannot calculate text dates in this manner in a Word table.
If you are not using date fields, the macro approach is still the most sensible option here.