If \(\rm \vec A=(3z^2+6y)\hat i-14 yz\hat j+20xz^2\hat k\), then the line integral \(\rm \int_C\vec A. d\bar r\) from (0, 0, 0) to (1, 1, 1), along the curve C; x = t, y = t2, z = t3 is:

1
\(\frac{31}{7}\)
2
6
3
5
4
4

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