this is the problem I've been thinking on for a while:
Let $f$ be a functional on $\mathbb{R}^3$
we know that $f(1,2,-1) = 2$ and $kerf = \{(x_1,x_2,x_3) \in \mathbb{R}^3 : 2x_1-x_2+3x_3 = 0\}$
I was able to do these steps: $base \ kerf = ((1,2,0),(0,1,3))$
$f(x,y,z) = ax + by + cz \ for \ a,b,c\in\mathbb{R}$
$f(1,2,-1) = 2 \implies c = a+2b-2$
I don't really understand how to connect these facts
I think the second element in the base for the kernel should be $(0,3,1)$. Well, you just need to find the images of the standard basis of $\mathbb{R^3}$, and you know what $f$ is. For example:
$2=f(1,2,-1)=f(1,2,0)-f(0,0,1)=-f(0,0,1)$
And so $f(0,0,1)=-2$. Next:
$0=f(0,3,1)=3f(0,1,0)+f(0,0,1)=3f(0,1,0)-2$
Which gives you $f(0,1,0)=\frac{2}{3}$. I'll leave you to compute $f(1,0,0)$.