The answer lies in the shadows .....

I'll give a general solution for respective projected areas $A, B$ and $C$ first.

Let $u$ and $v$ be vectors representing (any) two sides of $T$ in both direction and magnitude. The area of $T$ is then one half the magnitude of the vector cross-product of these two vectors, i.e., $(\frac{1}{2}) |u \times v|$ .

With $i,j,k$ being the standard unit vectors, we are thus given that

(i) $(\frac{1}{2}) |(u \times v) \cdot k| = A$ ,

(ii) $(\frac{1}{2}) |(u \times v) \cdot j| = B$ , and

(iii) $(\frac{1}{2}) |(u \times v) \cdot i| = C$ .

By Pythagoras we then have that $(\frac{1}{2}) |u \times v| = \sqrt{A^{2} + B^{2} + C^{2}}$ .

In this case, then, the area of $T$ is $\sqrt{4 + 36 + 81} = \boxed{11}$ .