MathematicsMedium202×since 2002Q5861Let a→=6i^+j^−k^\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}a=6i^+j^−k^ and b→=i^+j^\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}b=i^+j^. If c→\overrightarrow{\mathrm{c}}c is a is vector such that ∣c→∣≥6,a→⋅c→=6∣c→∣,∣c→−a→∣=22|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}∣c∣≥6,a⋅c=6∣c∣,∣c−a∣=22 and the angle between a⃗×b⃗\vec{a} \times \vec{b}a×b and c⃗\vec{c}c is 60∘60^{\circ}60∘, then ∣(a⃗×b⃗)×c⃗∣|(\vec{a} \times \vec{b}) \times \vec{c}|∣(a×b)×c∣ is equal to:A326\frac{3}{2} \sqrt{6}236B92(6−6)\frac{9}{2}(6-\sqrt{6})29(6−6)C92(6+6)\frac{9}{2}(6+\sqrt{6})29(6+6)D323\frac{3}{2} \sqrt{3}233Check answerSkip