I have to find equation of a plane passing through a point $M(3,1,2,1)$ and a line $l$ which is intersection of the planes $\alpha$ and $\beta$.
The elements are in the extended euclidean space $E_3^*$
$$ l: \left\{ \begin{array}{c} \alpha: x+y=0 \\ \beta: y+z-t=0 \\ \end{array} \right. $$
OK, my reasoning is the following. I have to find two points passing through the line $l$, for example $N(1, -1, 1, 0)$ and $P(-2, 2, -1, 1)$ I can form a determinant
$$ \begin{array}{|cccc|} x & y &z & t & \\ 3 & 1 &2 & 1 &\\ 1& -1 & 1 & 0 &\\ -2 & 2 & -1 & 1\\ \end{array} $$
which when solved will give the equation of the desired plane.
Is this reasoning correct? Also, is this like using the fact that the vector product of 3 point points is 0 when they are coplanar?
Yes, your reasoning is correct.
Another approach is this:
A plane through the intersection line of planes $\alpha$ and $\beta$ belongs to the pencil of planes defined by $\alpha$ and $\beta$, and a plane in this pencil has equation $$\lambda(x+y)+\mu(y+z-t)=0$$ for some $\lambda,\mu$ not both $0$ (actually $(\lambda,\mu)$ are unique up to a non-zero factor, i.e. they define a point in $\mathbf P^1(\mathbf R)$.
Then, a plane in the pencil passes through $M(3,1,2,1)$ if and only if $$\lambda(3+1)+\mu(1+2-1)=0,$$ whence the ration $\lambda/\mu$.