True or false :

Let $\displaystyle T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation defined by $\displaystyle T(1,0,0)=(1,1,0)$, $\displaystyle T(0,1,0)=(-2,1,1)$, $\displaystyle T(0,0,1)=(0,3,-2)$.

If $\displaystyle C$ is the cube formed by the vertices $\displaystyle (0,0,0)$, $\displaystyle (1,0,0)$, $\displaystyle (0,1,0)$, $\displaystyle (0,0,1)$, then $\displaystyle T(C)=3^2$.

My attempt : By intuition its false because $\displaystyle 3^2$ seems more like an area, but it could be a volume.

Ok, so I have formed $\displaystyle [T]_c^{c}$ in order to find out that $\displaystyle T(x,y,z)=(x-2y,x+y+3z,y-2z)$. Then I found the vertices passed by the transformation to be $\displaystyle (0,0,0)$, $\displaystyle (1,1,0)$, $\displaystyle (-2,1,1)$ and $\displaystyle (0,3,-2)$. Now it remains to calculate the volume of the solid formed by these vertices. I'm not really good when it comes to draw 3 dimensional solids. Especially this "prism" looking volume. I've no idea how to find out its volume.

Oh.... let me see, I remember I've read in a calculus III book that I must calculate a jacobian matrix and check out the determinant, or something like that. If it's 1 then the volume of the cube is the same as the volume of this prism. And if it's 2 then the volume of the prism is half or twice the volume of the cube, etc. I don't know which jacobian to find though.