Let number of red only balls be \(a\) and green only ball be \(b\) and both red and green be \(c\).

So we can say \(a+b+c=87\).

2/7 of the balls that have red color also have green color

So \(\frac{c}{(a+c)}= \frac{2}{7}\).

3/7 of the balls that have green color also have red color

So \(\frac{c}{(b+c)}=\frac{3}{7}\).

Inverting both fractions and adding them

\(\frac{(a+c)}{c}+\frac{(b+c)}{c}=\frac{7}{3}+\frac{7}{2}\)

Subtracting 1 from both sides

\(\frac{a+b+2c}{c}-1= \frac{35}{6}-1\).

\(\frac{a+b+c}{c}=\frac{29}{6}\)

Inverting again

\(\frac{c}{a+b+c}=\frac{6}{29}\).

Our objective was always to find \(\frac{c}{a+b+c}\).

Hence option D is correct.