Equation of line passing through i^+3j^+2k^ and ⊥ to the line r→=(i^+2j^−k^)+λ(2i^+j^+k^) and r→=(2i^+6j^+k^) +μ(i^+2j^+3k^) is
r→=(i^+2j^−k^)+λ(−i^+5j^−3k^)
r→=i^+3j^+2k^+λ(i^−5j^+3k^)
r→=i^+3j^+2k^+λ(i^+5j^+3k^)
r→=i^+3j^+2k^+λ(−i^−5j^−3k^)
The required line passes through the point i^+3j^+2k^ and is perpendicular to the lines r→=(i^+2j^−k^)+λ(2i^+j^+k^) and
r→=(2i^+6j^+k^) +μ(i^+2j^+3k^)
therefore, it is parallel to the vector b→=(2i^+j^+k^)×(i^+2j^+3k^)=(i^−5j^+3k^)
Hence, the equation of the required line is r→=(i^+3j^+2k^)+λ(i^−5j^+3k^)