A 3d line intersecting 2 other 3d lines The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$  and intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
My solution is as follow
$\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4}$ represent the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$  and intersecting the line and intersecting the line ${L_1}:9x + y + z + 4 = 0 = 5x + y + 3z$ & ${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
Let $\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4} = {\lambda _i}$  where  $\left( {x,y,z} \right) = \left( {a + 2{\lambda _i},b + 3{\lambda _i},c + 4{\lambda _i}} \right)$
$ \Rightarrow 9a + 18{\lambda _1} + b + 3{\lambda _1} + c + 4{\lambda _1} + 4 = 0 \Rightarrow \frac{{9a + b + c + 4}}{{ - 25}} = {\lambda _1}$ after putting the values in $L_1$
$5a + 10{\lambda _1} + b + 3{\lambda _1} + 3c + 12{\lambda _1} = 0 \Rightarrow \frac{{5a + b + 3c}}{{ - 25}} = {\lambda _1}$
$9a + b + c + 4 = 5a + b + 3c \Rightarrow 4a - 2c + 4 = 0$
${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
$\left( {x,y,z} \right) = \left( {a + 2{\lambda _2},b + 3{\lambda _2},c + 4{\lambda _2}} \right)$
$a + 2{\lambda _2} + 2b + 6{\lambda _2} - 3c - 12{\lambda _2} - 3 = 0 = 2a + 4{\lambda _2} - 5b - 15{\lambda _2} + 3c + 12{\lambda _2} + 3$
$ \Rightarrow a + 2b - 3c - 3 - 4{\lambda _2} = 0 = 2a - 5b + 3c + 3 + {\lambda _2} \Rightarrow \frac{{a + 2b - 3c - 3}}{4} = {\lambda _2} = \frac{{2a - 5b + 3c + 3}}{{ - 1}}$
$ \Rightarrow a + 2b - 3c - 3 =  - 8a + 20b - 12c - 12 \Rightarrow 9a - 18b + 9c + 9 = 0 \Rightarrow a - 2b + c + 1 = 0$
$ \Rightarrow a - 2b + c + 1 = 0\& 2a - c + 2 = 0$
Let $b =  - 1 \Rightarrow a + c + 3 = 0\& 2a - c + 2 = 0 \Rightarrow c =  - \frac{4}{3}\& a =  - \frac{5}{3}$
$\frac{{x + \frac{5}{3}}}{2} = \frac{{y + 1}}{3} = \frac{{z + \frac{4}{3}}}{4} = \lambda  \Rightarrow \left( {x,y,z} \right) = \left( { - \frac{5}{3}\hat i - \hat j - \frac{4}{3}\hat k} \right) + \lambda \left( {2\hat i + 3\hat j + 4\hat k} \right)$
But my answer is not matching. Can you tell me the error that I have made. Each steps is elaborated

To find a line parametric equation, we have to find a point on the line, and then use the cross product of the normal vectors of the the two planes given.
The two planes for the first line are
$9x + y + z = -4 $
$5 x + y + 3 z = 0 $
To find a point on both these planes, set $x = 0$ and solve the system
$ y + z = -4 $
$ y + 3 z = 0 $
The solution is $y = -6, z = 2 $ so our point is $(0, -6, 2) $
Next the direction vector of the line is along the cross product of the first normal and second normal
$d = (9, 1, 1) \times (5, 1, 3) = (2, -22, 4) = 2 (1, -11, 2) $
So the parametric equation of the first line is
$ P_1(t) = (0, -6, 2) + t (1, -11, 2) $
Similarly, for the second line, the planes are
$ x + 2 y - 3 z = 3 $
$ 2 x - 5 y + 3 z = -3 $
To find a point of the line, it must satisfy both equations, so set $y = 0$ and solve the system
$ x - 3 z = 3 $
$ 2 x + 3 z = -3 $
The solution is $x = 0 , z = -1 $, so our point is $(0, 0, -1) $
And the direction vector is
$ (1, 2, -3) \times (2, -5, 3) = -9 (1, 1, 1) $
So the parametric equation of the second line is
$ P_2 (s)= (0, 0, -1) + s (1, 1, 1) $
Now, since the line whose equation is requested is parallel to $(2,3,4)$ and it intersects both the first and second lines obtained above, then the setup is as follows:
$ P_1(t) - P_2(s) = \lambda (2, 3, 4) $
For some $t, s, \lambda$.  Writing the coordinates as column vectors, this last equation becomes the linear system
$ \begin{bmatrix} 1 && -1 && -2 \\ -11 && -1 && -3 \\ 2 && -1 && - 4 \end{bmatrix} \begin{bmatrix} t \\ s\\ \lambda \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ -3 \end{bmatrix} $
Its solution is $ (t, s, \lambda ) = ( - \dfrac{3}{5} , - 3, \dfrac{6}{5} ) $
Thus the first point of intersection on the first line is
$ P_1^* = (0, -6, 2) +\left( -\dfrac{3}{5} \right)(1, -11, 2) = (-\dfrac{3}{5}, \dfrac{3}{5}, \dfrac{4}{5} ) $
And the intersection with the second line is
$P_2^* = (0, 0, -1) +(- 3) (1, 1, 1) = (-3, -3, -4) $
We can use either point for the parametric equation of our line.  Using the second point, we get
$ Q(t) = (-3, -3, -4) + t (2, 3, 4) $
