Let \(a_{1},\ a_{2},\ a_{3} ...., a_{49}\) be in \(A.P.\) such that \(\displaystyle \sum_{k = 0}^{12} a_{4k + 1} = 416\) and \(a_{9} + a_{43} = 66\). If \(a_{1}^{2} + a_{2}^{2} + ..... + a_{17}^{2} = 140\ m\), then \(m\) is equal to
1
\(34\)
2
\(33\)
3
\(66\)
4
\(68\)