Find the equation of the plane which parallel to the lines `vecr=hati+hatj+lamda(2hati+hatj+4hatk) and (x+1)/(-3)=(y-3)/2=(z+2)/(1)` and is passing through the point (0, 1, -1).

Explanation

The plane is parallel to the given lines, which are directed along vectors `veca=2hati+hatj+4hatk and vecb=-3hati+2hatj+1hatk`.
Then the plane is normal to vector `vecaxxvecb-=|{:(hati,,hatj,,hatk),(2,,1,,4),(-3,,2,,1):}|=-7hati-14hatj+7hatk`
Also, the plane passes through the point (0, 1, -1)
Therefore, the equation of the plane is `-7(x-0)-14(y-1)+7(z+1)=0 or 7x+14y-7z=21`