Show that the line
`vecr = (hati+hatj-hatk)+lambda(3hati-hatj)`
`vecr = (4hati-hatk)+mu(2hati+3hatk)` are coplanar. Also find the equation of plane in which these lines lie.

Explanation

Here `veca_(1)=4hati-hatk, vecb_(1)=2hati+3hatk`
`veca_(2) = hati+hatj-hatk, vecb_(2)=3hati-hatj`
Now, `veca_(2) - veca_(1) = -3hati+hatj`
`vecb_(1)xxvecb_(2)=|{:(hati,hatj,hatk),(2,0,3),(3,-1,0):}| = hati+9hatj-2hatk`
and `(veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2))`
`=(-3hati+hatj).(3hati+9hatj-2hatk)`
`= - 9+9 = 0`

Therefore the given lines are coplanar.
Now equation of plane containing these line is
`(vecr-veca_(1)).(vecb_(1)xxvecb_(2)) = 0`
`rArr vecr.(vecb_(1)xxvecb_(2)) = veca_(1).(vecb_(1) xx vecb_(2))`
`rArr vecr.(3hati+9hatj-2hatk)=(4hati-hatk).(3hati+9hatj-2hatk)`
`rArr vecr.(3hati+9hatj-2hatk) = 14`.