Suppose there is a linear translation $T: \mathbb{R^3} \rightarrow \mathbb{R^3}$ such that $$T(\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}) = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \ T(\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}) = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix},\ \text{and} \ T(\begin{bmatrix} 0 \\ 0 \\ 7 \end{bmatrix}) = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$$
1.) With this information, how do I find $$T(\begin{bmatrix} 4 \\ 4 \\ 15 \end{bmatrix})?$$
I'm not sure how to do the latex notation for the vectors, so I'll do it as ordered triples. Using linearity, you can figure out what your transformation does to the standard basis vectors. So from the third equation, we have $\frac 1 7 T((0,0,7))=T((0,0,1))=\frac 1 7 (1,1,0)=(\frac 1 7,\frac 1 7,0)$
Use that with the first equation to get $T((1,0,0))=T((1,0,1))-T((0,0,1))=(2,1,0)-(\frac 1 7,\frac 1 7,0)=(\frac {13} 7,\frac 6 7,0)$
Finally, rather than go all the way to the basis fector, I'll use use 2 copies of the second one to save a bit of math. So,
$T((4,4,15))=2T((1,2,3))+2T((1,0,0))+9T((0,0,1))=2(0,1,1)+2(\frac {13} 7,\frac 6 7,0)+9(\frac 1 7,\frac 1 7,0)=(5,5,2)$