<s>[INST] The equation of the line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ intersecting the lines $9x+y+z+4=0 =5x+y+3z$ & $x+2y-3z-3=0=2x-5y+3z+3$ The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$  intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$  &  $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$ is
(A)  $7x + y + 2z + 2 = 0 = x - 2y + z + 1$
(B)  $7x - y + 2z + 2 = 0 = x + 2y + z + 1$
(C) $x + 7y + z + 1 = 0 = 2x - y + 2z + 2$
(D) None of these
I am solving but only putting the general equation
Let the line be represented by  $\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4} \Rightarrow {L_m}:\overrightarrow r  = a\hat i + b\hat j + c\hat k + s\left( {2\hat i + 3\hat j + 4\hat k} \right) \Rightarrow {L_m}:\overrightarrow r  = \overrightarrow {{c_m}}  + s\overrightarrow {{t_m}} $
The equation of the line  $9x + y + z + 4 = 0 = 5x + y + 3z \Rightarrow {L_2}$
On calculating we get ${L_2}:\frac{x}{{ - 1}} = \frac{{y + 6}}{{11}} = \frac{{z - 2}}{{ - 2}} \Rightarrow \overrightarrow r  = 0\hat i - 6\hat j + 2\hat k + t\left( { - \hat i + 11\hat j - 2\hat k} \right) \Rightarrow {L_2}:\overrightarrow r  = \overrightarrow {{c_2}}  + t\overrightarrow {{t_2}} $
The equation of the line  $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3 = {L_3}$
On calculating we get ${L_3}:\frac{x}{1} = \frac{{y - 1}}{1} = \frac{{z - \frac{2}{3}}}{1} \Rightarrow \overrightarrow r  = 0\hat i + \hat j + \frac{2}{3}\hat k + u\left( {\hat i + \hat j + \hat k} \right) \Rightarrow {L_3}:\overrightarrow r  = \overrightarrow {{c_3}}  + u\overrightarrow {{t_3}} $
It is getting more and more complicated as need to find $\left( {\overrightarrow {{c_2}}  - \overrightarrow {{c_m}} } \right).\left( {\overrightarrow {{t_m}}  \times \overrightarrow {{t_2}} } \right) = 0$ & $\left( {\overrightarrow {{c_3}}  - \overrightarrow {{c_m}} } \right).\left( {\overrightarrow {{t_m}}  \times \overrightarrow {{t_3}} } \right) = 0$
Can this be solved by some other method
 [/INST] Answer D since none of the lines in A, B, C is parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}.$ Indeed, solving the pairs of associated homogeneous equations, we find that:

*

*(A) is directed by the vector $(-1,1,3)$

*(B) by $(1,1,-3)$

*(C) by $(1,0,-1).$
If the question was more open, I would have solved it the following way:
find $(a,b,c)\in L_3$ such that $(a,b,c)+t(2,3,4)\in L_2$ for some $t\in\mathbb R$ (write the 4 equations in the 4 unknowns $a,b,c,t$ and eliminate $t$ to get 3 equations in 3 unknowns). I found $(a,b,c)=(-3,-3,-4)$ (and $t=\frac65$).
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