Line intersecting 2 lines and parallel to another 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
 A: 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 $L$ lies on 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.
A: Let us use vector notation for the sake of simplicity:
$L_1: \vec r_1+t_1\vec l_1$ ,where $\vec l_1=(1,2,1)$ and $\vec r_1=(1,2,-2)$.
Likewise
$L_2: \vec r_2+t_2\vec l_1$ 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
$$ L: x=1.5+2t'', y=3+7t'', z=-1.5+3t''$$
A: The geometry of this question is interesting, most notably for the curious conditions governing the existence of a solution, so I would like to provide an answer that emphasizes this aspect, and in the process illustrate how the problem can be tackled in the most general case, not just for the particular example of this question.  Then, at the very end, the solution to this specific question will also be provided in a way that is hopefully as clear as it is simple.
In vector terms, we have
L1:$\ \ \vec{P} + t_1\vec{D_1}$
L2:$\ \ \vec{Q} + t_2\vec{D_2}$
L3:$\ \ \vec{R} + t_3\vec{D_3}$ 
and seek $\vec{S}$
L:$\ \ \vec{S} + t\vec{D_3}$
such that L intersects both L1 and L2.  
It is interesting to note that there are solutions when the line directions $\vec{D_1}$, $\vec{D_2}$ and $\vec{D_3}$ are not linearly independent.  
If $\vec{D_1}$, $\vec{D_2}$ and $\vec{D_3}$ are not linearly independent, then there is no solution except when L1 and L2 are located in a plane parallel to L3 without both individually being parallel to L3.  The solution then is $\vec{S} = \vec{P}$.
This case occurs if $\ ((\vec{P} - \vec{Q}) \times \vec{D_1}) \cdot \vec{D_3} = 0$ and $\ (\vec{D_3} \cdot \vec{D_1} ≠ 0\ $ or $\ \vec{D_3} \cdot \vec{D_2} ≠ 0)\ $.  
All other solutions involve linearly independent $\vec{D_1}$, $\vec{D_2}$ and $\vec{D_3}$, in which case the solution is
$\vec{S} = \vec{P} + t_1\vec{D_1} + t_3\vec{D_3} = \vec{Q} + t_2\vec{D_2}$
for some $t_1$, $t_2$ and $t_3$.
$\vec{S} = \vec{P} + t_1\vec{D_1}$ and $\vec{Q} + t_2\vec{D_2}$ are then the intersection points of L with L1 and L2 respectively.  
Setting  
$\vec{P}= \begin{pmatrix} p_x \cr p_y \cr p_z \cr \end{pmatrix},\ \ \vec{Q}= \begin{pmatrix} q_x \cr q_y \cr q_z \cr \end{pmatrix}$,  
$\vec{D_1}= \begin{pmatrix} x_1 \cr y_1 \cr z_1 \cr \end{pmatrix},\ \ \vec{D_2}= \begin{pmatrix} x_2 \cr y_2 \cr z_2 \cr \end{pmatrix},\ \ \vec{D_3}= \begin{pmatrix} x_3 \cr y_3 \cr z_3 \cr \end{pmatrix}$,  
the system of equations to solve is  
$p_x + t_1 x_1 + t_3 x_3 = q_x + t_2 x_2$
$p_y + t_1 y_1 + t_3 y_3 = q_y + t_2 y_2$
$p_y + t_1 z_1 + t_3 z_3 = q_z + t_2 z_2$.  
As an example, in this question we then have  
$t_1 - t_2 + 2t_3 = 1$
$2t_1 - 2t_2 + 7t_3 = -1$
$t_1 - 3t_2 + 3t_3 = 5$,  
giving $\ \ t_1 = {1 \over 2}$, $\ \ t_2 = -{5 \over 2}\ $ and $\ \ t_3 = -1$,
and so the two intersection points are  
$\vec{S} = \vec{P} + t_1\vec{D_1} = \begin{pmatrix} 1 \cr 2 \cr -2 \cr \end{pmatrix} + {1 \over 2} \begin{pmatrix} 1 \cr 2 \cr 1 \cr \end{pmatrix} = {1 \over 2} \begin{pmatrix} 3 \cr 6 \cr -3 \cr \end{pmatrix}= \begin{pmatrix} {3 \over 2}  \cr 3 \cr -{3 \over 2} \cr \end{pmatrix}$
and
$\vec{Q} + t_2\vec{D_2} = \begin{pmatrix} 2 \cr 1 \cr 3 \cr \end{pmatrix} - {5 \over 2} \begin{pmatrix} 1 \cr 2 \cr 3 \cr \end{pmatrix} = -{1 \over 2} \begin{pmatrix} 1 \cr 8 \cr 9 \cr \end{pmatrix}= \begin{pmatrix} -{1 \over 2}  \cr -4 \cr -{9 \over 2} \cr \end{pmatrix}$.
