I know there are similar questions here but I can't understand them. This one, for example Find the line that intersects two other lines
So, in my problem I have to find the line named $T$ that passes through the point $(1,1,0)$ and cuts line $$r: \left\{ \begin{array}{l} x+z=0 \\y=-2 \end{array} \right. \qquad \qquad s: \left\{ \begin{array}{l} x=1 \\y-z+3=0 \end{array} \right.$$. \ The thing is that I don't really understand what to do. I mean, the why of doing something.
On
The line to be found must be coplanar to the straights r and s and passing through $T(1,1,0)$. The equation of such a line, written in parametric form is as follows:
$\frac{x-1}{l}=\frac{y-1}{m}=\frac{z}{n}$.
For two straights to be coplanar, the determinant
$$\begin{pmatrix}x_{2}-x_{1}& y_{2}-y_{1}& z_{2}-z_{1}\\ l&m&m\\ l_{0}&m_{0}&n_{0}\end{pmatrix}$$,
must be null.
Substituting to $ l_{0},m_{0},n_{0}$,once the parameters of the line r, another time those of the line s,we obtain two equations in $l,m,n$, which solved as a function of $n$, give as a solution:
$l=0,m=3,n=1$.
The equation of our goal becomes:
$x=1, y=3z+1$.
Hint:
Denote $A$ the point with coordinates $(1,1,0)$. This point, together with $r$ determines a plane $\Pi$ belonging to the pencil of planes with basis the planes $x+z=0$, $y+2=0$, i.e. its equation is $$\lambda(x+z)+\mu(y+2)=0,\quad \lambda,\mu\in\mathbf R.$$ Note that these coefficients are defined up to a nonzero factor.
Determine the coefficients $\lambda,\mu$. If $B$ is the intersection of line $s$ with $\Pi$, the sought line is $(AB)$.
Variant: Determine in the same way the equation of the plane $\Pi'$ defined by $A$ and the line $s$. The sought line in the intersection of $\Pi$ and $\Pi'$.