If \( a=i+j+k, b=i+j, c=i \) and \( (a \times b) \times c=\lambda a...
If \( a=i+j+k, b=i+j, c=i \) and \( (a \times b) \times c=\lambda a+\mu b \)
\( \mathrm{P} \) then \( \lambda+\mu= \)
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