Let ΔPQR be a triangle. Let 𝐚 = QR, 𝐛 = RP. If |𝐚| = 12, |𝐛| = 4√3 and 𝐛.𝐜 = 24, then which of the following is (are) true?

Using vectors summation method, −𝐚 = 𝐛 + 𝐜
|𝐚|² = |𝐜|² + |𝐛|² ± 2|𝐜||𝐛|cosP
|𝐚|² = |𝐜|² + |𝐛|² + 2|𝐜||𝐛|cos(π − P)
|𝐚|² = |𝐜|² + |𝐛|² − 2𝐜 ⋅ 𝐛
|𝐚|² = 12² = 144
(4√3)² = 48
144 = |𝐜|² + 48 − 2(24)
144 = |𝐜|² + 48 − 48
|𝐜|² = 144
|𝐜| = 12
Therefore, |𝐜|² − |𝐚| = 144 − 12 = 132 ≠ 12
|𝐜|² + |𝐚| = 144 + 12 = 156 ≠ 30
Since −𝐚 = 𝐛 + 𝐜, then 𝐚 = −𝐛 − 𝐜
𝐚 ⋅ 𝐛 = (−𝐛 − 𝐜) ⋅ 𝐛 = −|𝐛|² − 𝐜 ⋅ 𝐛 = −48 − 24 = −72 ≠ 0
|𝐚 × 𝐛 + 𝐜 × 𝐚| = |𝐚 × 𝐛 − 𝐚 × 𝐜| = |𝐚 × (𝐛 − 𝐜)| = |𝐚||𝐛 − 𝐜|sinθ
Since −𝐚 = 𝐛 + 𝐜, then 𝐜 = −𝐚 − 𝐛
|𝐜| = 12
|𝐜|² − |𝐚| = 144 − 12 = 132
Option D is incorrect.
Let's check option A:
|𝐚 × 𝐛 + 𝐜 × 𝐚| = |𝐚 × 𝐛 − 𝐚 × 𝐜| = |𝐚 × (𝐛 − 𝐜)| = |𝐚||𝐛 − 𝐜|sinθ
|𝐚 × 𝐛| = |𝐚||𝐛|sin(π − P) = |𝐚||𝐛|sinP
|𝐚 × 𝐛 + 𝐜 × 𝐚| = |𝐚 × (𝐛 − 𝐜)| = |𝐚||𝐛 − 𝐜|sinθ
|𝐚| = 12, |𝐛| = 4√3, 𝐜 ⋅ 𝐛 = 24
|𝐜|² = 144
|𝐜| = 12
|𝐜|² − |𝐚| = 144 − 12 = 132
Only option D is true. However, the calculations show that none of the options are true.