Let a=^i+4^j+2^k, b=3^i−2^j+7^k and c=2^i−^j+4^k. Find a vector p which is perpendicular to both a and b and p⋅c=18.

a=^i+4^j+2^k, b=3^i−2^j+7^k
Since we need a vector perpendicular to both a and b, it has to be parallel to a×b
a×b=(^i+4^j+2^k)×(3^i−2^j+7^k)=−2^k−14^j+32^i+6^j+4^i=32^i−1^j−2^k
The unit vector in this direction can be written as
32^i−^j−2^k/√1024+1+4=32^i−^j−2^k/√1029
Since p⋅c=18, ∴k√1029×(32^i−^j−2^k)⋅(2^i−^j+4^k)=18
∴k√1029×(64+1−8)=18
∴k=18√1029/57
∴p=k/√1029×(32^i−^j−2^k)=18/57×(32^i−^j−2^k)=6/19×(32^i−^j−2^k)=384/19^i−6/19^j−12/19^k