Finding a linear map $T:\mathbb R^3 \to \mathbb R^3$ that satisfy conditions about the kernel and image Finding a linear map $T:\mathbb R^3 \to \mathbb R^3$ that satisfy conditions about the kernel and image
  
  1)find linear map $T:\mathbb R^3 \to \mathbb R^3$ so that $\operatorname{Im} (T)=\{(x,y,z)|2x-y+z=0\}$
  
  2)find linear map $T:\mathbb R^3 \to \mathbb R^3$ so that $\operatorname{Ker} (T)=\{(x,y,z)|2x-y+z=0\}$


I choose for (1) : $T(x,y,z)=(x+y+z,3x+3x+3z,x+y+z)$

and for (2):$T(x,y,z)=(2x-y+z,4x-2y-2z,6x-3y+3z)$

so Does this answer correct and Is there any formal way to find $T$?

thanks
(1) Your answer is wrong. The image of your map $T$ is$$\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,2x-y+z=0\wedge x=z\right\}.$$You can solve the problem by finding a basis $\{e,f\}$ of $\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,2x-y+z=0\right\}$ (for instance $e=(1,2,0)$ and $f=(0,1,1)$) and define $T$ such that $T(e_1)=T(e_2)=e$ and $T(e_3)=f$, where $\{e_1,e_2,e_3\}$ is the standard basis of $\mathbb{R}^3$.

(2) You were almost right. The second $-2$ shouldn't have a minus sign in it. It should have been$$T(x,y,z)=(2x-y+z,4x-2y+2z,6x-3y+3z).$$
(1) Your answer is wrong. The image of your map $T$ is$$\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,2x-y+z=0\wedge x=z\right\}.$$You can solve the problem by finding a basis $\{e,f\}$ of $\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,2x-y+z=0\right\}$ (for instance $e=(1,2,0)$ and $f=(0,1,1)$) and define $T$ such that $T(e_1)=T(e_2)=e$ and $T(e_3)=f$, where $\{e_1,e_2,e_3\}$ is the standard basis of $\mathbb{R}^3$.

(2) You were almost right. The second $-2$ shouldn't have a minus sign in it. It should have been$$T(x,y,z)=(2x-y+z,4x-2y+2z,6x-3y+3z).$$