Use standard series to show that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 2 r - p ) = \frac { 1 } { 6 } n ( n + 1 ) \left( 3 n ^ { 2 } + ( 3 - 2 p ) n - p \right) ,$$
where \(p\) is a constant.
Given that the coefficients of \(n ^ { 3 }\) and \(n ^ { 4 }\) in the expression for \(\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 2 r - p )\) are equal, find the value of \(p\).