Show that there is one linear mapping $\varphi : \mathbb{Q}^3 \rightarrow \mathbb{Q}^3$

As $\{v_{1},v_{2},v_{3}\}$ is a linearly independent set of $3$ vectors , they form a basis for $\mathbb{Q}^{3}$.

So you only have to check whether $\{\phi(v_{1}),\phi(v_{2}),\phi(v_{3})\}$ form a basis for $\mathbb{Q^{3}}$ in order to figure out if it's an isomorphism or not. If so then you will have a function mapping basis to basis which will extend to a linear map which will be a bijection and hence an isomorphism.

In this case $\phi(v_{3})=0$ . So you do not have an isomorphism.

However since any vector $v=(x,y,z)$ in $\mathbb{Q^{3}}$ can be written as

$v=c_{1}v_{1}+c_{2}v_{2}+c_{3}v_{3}$.

If you define $\phi(v)=\sum_{i=1}^{3}c_{i}\phi(v_{i})$ you will get a unique linear map $\phi$.

To express $\phi$ as an explicit function, try and find a solution for variables $c_{1},c_{2},c_{3}$ such that

$(x,y,z)=(c_{1}+2c_{2}-c_{3},c_{1}+c_{2}+c_{3},c_{1}+5c_{2}-c_{3})$ (this is the only part that requires a bit of labour. it is not tough).

Then you will be able to write

$(x,y,z)=c_{1}v_{1}+c_{2}v_{2}+c_{3}v_{3}$

And finally apply $\phi$ on both sides to get the explicit formula for the transformation.

$\phi(x,y,z)=c_{1}(1,1,-1)+c_{2}(2,1,-5)$