The correct option is C −1
Since position vector of the point with respect to origin will be:
→r=2^i−6^j−12^k
Equation of torque for a force →F acting at a point whose position vector is →r about a reference point
→τ=→r×→F=∣∣
∣
∣∣^i^j^k2−6−12α36∣∣
∣
∣∣
→τ=^i[(−6×6)−(−12×3)]−^j[(2×6)−(−12×α)]+^k[(2×3)−(−6×α)]
⇒→τ=0^i−(12+12α)^j+(6+6α)^k
⇒→τ=0^i−(12+12α)^j+(6+6α)^k
For angular momentum to be conserved about origin,
∴τext=0 about origin
⇒0^i−(12+12α)^j+(6+6α)^k=0
Equating the coefficients of ^i, ^j, ^k respectively to Zero, we get:
⇒12+12α=0 and 6+6α=0
∴α=−1212=−66=−1