determinant of a a linear operator I understand that matrix of $T$ will be $6\times 6$ and will consists of $1,0$ but not able to find out what will be the determimant.
Here the space of matrices is viewed as a 6-dimensional space.  The 2 by 3 matrix structure is irrelevant.  Therefore it is preferable to view this map as $T:\mathbb{R}^6\to \mathbb{R}^6$ given by $(x_1,x_2,x_3,x_4,x_5,x_6)\mapsto(x_6,x_4,x_1,x_3,x_5,x_2)$.  Then one immediately recongnizes a permutation matrix.  Hence its determinant is either $1$ or $-1$.  Now all you have to do is determine whether the permutation is even or odd, for example by decomposing it into cycles.

I just noticed the minus sign on $x_6$, so this is not exactly a permutation matrix.  That changes the sign of the determinant to its opposite.
I think that for the transformation to be defined, $T$ must be a $2x2$ matrix.After which you will have a system of equations. Then you may solve for the entries of $T$.