Easier way to calculate Nullity of triple composition of a linear transformation?

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$T:R^3\to R^3$

$T(x,y,z)=(x+3y+2z,3x+4y+z,2x+y-z)$

I want to find the dimension of the Null Space of $T^3$

Here Nullity of $T$ is not $0$, so I can't say anything directly about the nullity of $T^3$

Normally for the nullity of T, I would form the matrix of transformation by take the image of the standard basis of $R^3$, then row reduce to calculate the rank and use the dimension theorem.

But doing the same for $T^3$ is turning out to be hefty task and I am looking for a easier way.