If the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}$ intersect the $xy$ and $yz$ plane at points $A$ and $B$ respectively. If the volume of the tetrahedron $OABC$ is $V$ cubic units (where, $O$ is the origin) and point $C$ is $\left(\mathrm{1,0},4\right)$, then the value of $102V$ is equal to

On the $xy$ plane, $z=0$
$\Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=-1\Rightarrow x=-1,y=-1$
$\Rightarrow$ Coordinates of point $A$ are $\left(-1,-\mathrm{1,0}\right)$
On the $yz$ plane, $x=0$
$\Rightarrow \frac{-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}\Rightarrow y=\frac{1}{2},z=2$
$\Rightarrow$ Coordinates of point $B$ are $\left(0,\frac{1}{2},2\right)$
$\overrightarrow{OA}=-\widehat{i}-\widehat{j}$
$\overrightarrow{OB}=\frac{1}{2}\widehat{j}+2\widehat{k}$
$\overrightarrow{OC}=\widehat{i}+4\widehat{k}$
The volume of the tetrahedron $OABC$ is $V=\frac{1}{6}\left[\overrightarrow{OA} \overrightarrow{OB} \overrightarrow{OC}\right]$
$=\frac{1}{6}\left|\begin{array}{ccc}-1& -1& 0\\ 0& \frac{1}{2}& 2\\ 1& 0& 4\end{array}\right|$
$=\left|\frac{1}{6}\left[-1\left(2\right)+1\left(-2\right)\right]\right|=\frac{4}{6}=\frac{2}{3}$ cubic units

Hence, $102V=68$