Answer
Position vector can be written as
r=xi^+yi^+zi^
=(x0+at2)i^+bt3j^+ctk^
Now acceleration can be written as
a=dt2d2r=(2a)i^+6btj^
Now cross product
r×F=r×(ma)
=m(r×a)
=∣∣i^(x0+at2)2aj^bt36btk^ct0∣∣
=−6bct2i^−2actj^+(6x0bt+4abt3)k^
This means
r×F=0
In this specific case we can rewrite as
dL/dt=r×F=τ
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