Show that the vectors `hati-3hati+2hatk,2hati-4hatj-hatk and 3hati+2hatj-hatk` and linearly independent.

Let `alpha=hati-3hatj+2hatk`
`beta=2hati-4hatj-hatk`
and `gamma=3hati+2hatj-hatk`
Also, let `x alpha+ybeta+zgamma=0`
`thereforex(hati-3hatj+2hatk)+y(2hati-4hatj-hatk)+z(3hati+2hatj-hatk)=0`
or `(x+2y+3z)hati+(-3x-4y+2z)hatj+(2x-y-z)hatk=0`
Equating the coefficient of `hati,hatj and hatk`, we get
`x+2y+3z=0`
`-3x-4y+2z=0`
`2x-y-z=0`
Now, `|(1,2,3),(-3,-4,2),(2,-1,-1)|=1(4+2)-2(3-4)+3(3+8)=41 ne0`
`therefore` The above syste of equations have only trivial solution. thus, x=y=z=0
Hence, the vectors `alpha,beta and gamma` are linearly independet.
Trick consider the determinant of coefficients of `hati,hatj and hatk`
i.e., `|(1,-3,2),(2,-4,-1),(3,2,-1)|=1(4+2)+3(-2+3)+2(4+12)`
`=6+3+32=41ne0`
`therefore`The given vectors are non-coplanar. hence, the vectors are linearly independent.