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Solution :

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