Let we have the projective space $\mathbb{P}^3$. We have to calculate the projective plane which includes the projective line $\mathbb{S}:[ x_1-x_2+x_3=0=x_2-2x_3+x_4$ and the projective point $P=(1:2:0:1)$.
What I have done is:
First of all I have defined $\mathbb{S}=\mathbb{P}(W)$ and I have calculated the subspace $W$ solving the equations: $W=<(1,1,0,-1),(-1,0,1,2)>$
And now, I don't know if what I have done is ok:
I have calculated the plane (which I have called $T$) which includes the subspace $W$ and the point $(1,2,0,1)$ in the vector space; I mean, $T=(1,2,0,1)+<(1,1,0,-1),(-1,0,1,2)>$ and calculating the implicit equations: $T:[x_1-x_2+x_3+1=0, x_2-2x_3+x_4-3=0$
$T=${$(x_1,x_2,x_3,x_4)\in\mathbb{R^4}|x_1-x_2+x_3+1=0, x_2-2x_3+x_4-3=0$}
And now, if I'm not wrong, what the exercise asks to us is $\mathbb{S'}=\mathbb{P}(T)$ and I don't know if I'm doing it well but I've put that:
$\mathbb{S'}=\mathbb{P}(T)=${$(x_1:x_2:x_3:x_4)\in\mathbb{P}^4|x_1-x_2+x_3+1=0, x_2-2x_3+x_4-3=0$}
But I'm pretty sure that's wrong, owing to the fact that the equations of a projective subspace must be homogeneous.
I don't really know what I'm doing wrong and I neither know how to do it in a correct way.