Line through two given skew lines and origin Question
Find the direction cosines of the line through the origin which intersects each of the lines
$\displaystyle \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and  $\displaystyle \frac{x+2}{4} = \frac{y-3}{3} = \frac{z-4}{2}$.
My Attempt
Projection of a vector connecting the two lines on the perpendicular from the required line to the given two lines is zero. Hence
\begin{equation}
\begin{vmatrix}
1&2&3\\
2&3&4\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}

\begin{equation}
\begin{vmatrix}
-2&3&4\\
4&3&2\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}
This gives me two equations
\begin{equation}
a-2b+c=0
\end{equation}
\begin{equation}
3a-10b+9c=0
\end{equation}
I get two different ratios for the direction cosines (a,b,c) as
\begin{equation}
(5,3,1),(2,3,1)
\end{equation}
Both these direction cosines give me a line intersecting only one given line.
Is my approach wrong or does this mean that there is no line that intersects both given lines and also goes through the origin?
Highly appreciate your help on this.Thanks in advance.

You are correct that there is no such line other than a mistake you made in calculation. Here is another approach I used to validate.
An intersection point on line $1$ is $P (2s+1, 3s+2, 4s+3)$ and on line $2$, $Q (4t-2, 3t + 3, 2t + 4)$.
Now as the the intersecting line passes through origin $O$, we must have, $\vec {OP} = \lambda \ \vec {OQ}$ where $\lambda$ is a scalar constant and we must have $\lambda \ne 0$.
So, $(2s+1, 3s+2, 4s+3) = \lambda (4t-2, 3t + 3, 2t + 4)$
From first two, $s = \cfrac{18 \lambda - 5}{6}, t = \cfrac{12\lambda-1}{6\lambda}$
Plugging $\lambda$ into the third component and solving, $\lambda = 0$. Hence there is no such line that intersects both lines and passes through the origin.
