The correct option is A −1
Given,
→F=α^i+3^j+6^k..... (1)
→r=2^i−6^j−12^k ....(2)
We know for angular momentum to remain conserved, torque of force has to be zero.
⇒→τ=0
⇒→r×→F=0 [→τ=→r×→F]
⇒∣∣
∣
∣∣^i^j^k2−6−12α36∣∣
∣
∣∣=0
⇒^i(−36+36)−^j(12+12α)+^k(6+6α)=0
⇒12+12α=0 or 6+6α=0
⇒α=−1
The value of α is −1 for which angular momentum about origin is conserved.