Let the first term of the geometric progression be 216 and the common ratio be \(r\).

The fourth term is given by \(216r^3 = 64\).

Solving for \(r\), we have:

\(r^3 = \frac{64}{216} = \frac{8}{27}\)

\(r = \frac{2}{3}\)

The second term of the geometric progression is \(216r = 144\).

The third term of the geometric progression is \(216r^2 = 96\).

Let the first term of the arithmetic progression be \(a\) and the common difference be \(d\).

The second term of the arithmetic progression is \(a + d = 144\).

The fifth term of the arithmetic progression is \(a + 4d = 96\).

Solving the equations simultaneously:

\(a + d = 144\)

\(a + 4d = 96\)

Subtract the first equation from the second:

\(3d = -48\)

\(d = -16\)

Substitute \(d = -16\) into \(a + d = 144\):

\(a - 16 = 144\)

\(a = 160\)

The sum of the first 21 terms of the arithmetic progression is given by:

\(S_{21} = \frac{21}{2} (2a + 20d)\)

\(S_{21} = \frac{21}{2} (320 + 20(-16))\)

\(S_{21} = \frac{21}{2} (320 - 320)\)

\(S_{21} = 0\)