Htykuut

The problem is :

Find parametric equations of the line L intersecting the given lines L1 and L2 and parallel to the given line L3.

L1: x = 1 + (t1), y = 2 + 2(t1), z = -2 + (t1)

L2: x = 2 + (t2), y = 1 + 2(t2), z = 3 + 3(t2)

L3: x = 1 + 2(t3), y = 1 + 7(t3), z = 1 + 3(t3)

(t1), (t2) and (t3) are in R

Thank you

Let us use vector notation for the sake of simplicity:
$L_1: vec r_1+t_1vec l_1$ ,where $vec l_1=(1,2,1)$ and $vec r_1=(1,2,-2)$.
Likewise
$L_2: vec r_2+t_2vec l_1$ where $vec l_2=(1,2,3)$ and $vec r_2=(2,1,3)$. Let us call the line $L$ that intersects $L_1$ and $L_2$ and is parallel to $L_3$. So we get
$L: vec r+t vec l_3$ ,where $vec l_3=(2,7,3)$ and $vec r$ yet to be determined.

From intersection of $L_1$ and $L$ we get:
$$vec r_1+t_1 vec l_1=vec r+ t'_1 vec l_3$$
And from intersection of $L_2$ and $L$ we get:
$$vec r_2+t_2 vec l_2=vec r+ t'_2 vec l_3$$
If we subtract the last two vector equation we get:
$$vec r_1-vec r_2+t_1 vec l_1-t_2 vec l_2=(t'_1-t'_2)vec l_3$$
If we substitute the numerical values we get three equations as below:
$$left{begin{array}{c}-1+t_1-t_2=2(t'_1-t'_2) \1+2(t_1-t_2)=7(t'_1-t'_2)\-5+t_1-3t_2=3(t'_1-t'_2)end{array}right.$$
We can consider $t_1$ , $t_2$ and $t'_1-t'_2$ as three unknowns in the system of three equations to get: $t_1=0.5$, $t_2=-2.5$, and $t'_1-t'_2=1$.
So from intersecting $L$ and $L_1$ we can get $vec r$ as below
$$vec r=vec r_1+t_1 vec l_1-t'_1 vec l_3=(1,2,-2)+0.5(1,2,1)-t'_1 vec l_3=(1.5,3,-1.5)-t'_1 vec l_3$$
Then the parametric equation for $L$ will be:
$$L: vec r+t vec l_3=(1.5,3,-1.5)-t'_1 vec l_3+t vec l_3$$
or by having $t''=t-t'$ we get
$$L: vec r+t vec l_3=(1.5,3,-1.5)+t'' vec l_3$$
or
$$L: vec r+t vec l_3=(1.5,3,-1.5)+t'' (2,7,3)$$
or
$$ L: x=1.5+2t'', y=3+7t'', z=-1.5+3t''$$

If two lines intersect, then they are coplanar. In particular, $L$ and $L_1$ must lie on the same plane, as must $L$ and $L_2$, therefore $L$ is the intersection of these two planes. The cross product of the direction vectors of each of these pairs of lines is normal to the plane in which they lie, and you can extract a known point on each of $L_1$ and $L_2$ from their parametric equations, so you can easily construct an equation for each of these planes. Once you have the two equations, finding their intersection is also straightforward. Indeed, you only need to find a point in the intersection since you already know what the direction vector of the line must be, but it’s worth going through the entire calculation to ensure that the planes do in fact intersect.

Specifically, we have for the first plane’s normal $(2,7,3)times(1,2,1)=(1,1,-3)$ and so an equation for it is $$x+y-3z=9.$$ For the second plane, we have the normal $(2,7,3)times(1,2,3)=(15,-3,-3)$, giving the equation $$5x-y-z=6.$$ These two normals are not parallel, so the planes do intersect in a line. I’ll leave the rest to you.