If the area of a parallelogram whose diagonals are represented by vectors $3 \hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ is $\frac{\sqrt{117}}{2}$ sq. units, then $\lambda=$
- A-1
- B-2
- C-3
- D-4
✓ Correct answer: D
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