JEE Main 202313 Apr, Evening shiftMathematicsHardVector Algebra184× since 2002Let for a triangle ABC\mathrm{ABC}ABC, AB→=−2i^+j^+3k^\overrightarrow{\mathrm{AB}}=-2 \hat{i}+\hat{j}+3 \hat{k}AB=−2i^+j^+3k^ CB→=αi^+βj^+γk^\overrightarrow{\mathrm{CB}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}CB=αi^+βj^+γk^ CA→=4i^+3j^+δk^\overrightarrow{\mathrm{CA}}=4 \hat{i}+3 \hat{j}+\delta \hat{k}CA=4i^+3j^+δk^ If δ>0\delta > 0δ>0 and the area of the triangle ABC\mathrm{ABC}ABC is 565 \sqrt{6}56, then CB→⋅CA→\overrightarrow{C B} \cdot \overrightarrow{C A}CB⋅CA is equal toAOption A. 60BOption B. 54COption C. 120DOption D. 108CheckSkip · show solution
