<s>[INST] Equation of the line passing through the intersection of two lines and is parallel to another line. The Question is :
Find the equation of the line through the intersection of the lines $3x+2y−8=0,5x−11y+1=0$ and parallel to the line $6x+13y=25$
Here is how I did it..
$L_1 = 3x + 2y -8 = 0$
$L_2 = 5x -11y +1 = 0$
$L_3= ?$
$L_4 = 6x + 13y -25= 0$
I found the point of intersection : $(-2, -1)$
Using formula: $L_1 + kL_2$ = 0
$(3x + 2y -8) + k(5x -11y +1)=0$
$(3(2) + 2(1) -88) + k(5(2) -11(1) +1) = 0$
$(6 +2 -8) + k(10 -11 +1) =0$
$8-8 + k(11-11) =0$
$0 +k(0) = 0$
What's wrong ?
 [/INST] This will blow your mind. Create vectors from the coefficients of the lines $$\begin{array}{rlrlrl}
L_1 & = \begin{pmatrix} 3\\2\\-8\end{pmatrix} &
L_2 & = \begin{pmatrix} 5\\-11\\1\end{pmatrix} &
L_4 & = \begin{pmatrix} 6\\13\\-25\end{pmatrix} \end{array} $$
The intersection point $P_3$ between $L_1$ and $L_2$ is found with a vector cross product
$$ P_3 = \begin{pmatrix} 3\\2\\-8\end{pmatrix} \times \begin{pmatrix} 5\\-11\\1\end{pmatrix}   = \begin{pmatrix} -86\\-43\\-43\end{pmatrix} $$ with coordinates $$(x,y) = \left( \frac{-86}{-43}, \frac{-43}{-43} \right) = (2,1)$$
The coefficients of a line parallel to $L_4$ are $L_3 = (6,13,c_3)$. To make this line pass through $P_3$ set
$$ P_3 \cdot L_3 = 0 $$
$$ \begin{pmatrix} -86\\-43\\-43\end{pmatrix} \cdot \begin{pmatrix} 6\\13\\c_3\end{pmatrix} =0 $$
$$ \left. -43 c_3 -1075 =0  \right\} c_3 = -25 $$
So line $L_3$ is $L_3 = (6,13,-25)$ with equation
$$ \begin{pmatrix} x\\y\\1\end{pmatrix} \cdot \begin{pmatrix} 6\\13\\-25\end{pmatrix} =0 
\\ 6x+13y-25 = 0 $$
</s>