JEE Advanced 2014 Paper 2

2x1+3x2+4x3=11
Possibilities are (0, 1, 2); (1, 3, 0); (2,1,1); (4, 1, 0)
∴ Required coefficients

= (1 × 7 × 66) + (4 × 35 × 1) + (6 × 7 × 12) + (1 × 7)
= 462 + 140 + 504 + 7 = 1113