The flaw is in your diagram. For your method to work, the point $A$ must lie on the projection of the line onto the plane, otherwise the vector $\vec{AR}$ will be oblique to the plane.
While drawing the figure you have assumed that the point $A$ is lying on the projection of $L$ onto $\Pi$ while it's actually not.
My answer is based on Lee Yiyuan's suggestion.
Rewrite the equation of line 1 as
$$\left(\begin{matrix}x\\y\\z\end{matrix}\right)
=\left(\begin{matrix}8\\-9\\-1\end{matrix}\right)
+\left(\begin{matrix}3\\-16\\-2\end{matrix}\right) t.
$$
A vector perpendicular to both lines is
$$\begin{vmatrix}i&j&k\\3&-16&-2\\3&8&-5\end{vmatrix}
=96i+9j+72k.$$
One of the vectors joining two lines is
$$\left(\begin{matrix}8\\-9\\-1\end{matrix}\right)
-\left(\begin{matrix}15\\29\\5\end{matrix}\right)
=\left(\begin{matrix}-7\\-38\\-6\end{matrix}\right).
$$
Calculate the projection of this vector to the vector perpendicular to both lines, and then take its absolute value as the final answer.
\begin{split}
& \mbox{Distance between $L_1$ and $L_2$}\\
=&\left\lvert\frac{\langle(-7,-38,-6),(96,9,72)\rangle}{\lVert(96,9,72)\rVert}\right\rvert\\
=&\left\lvert\frac{-1446}{\sqrt{14481}}\right\rvert\\
=&\frac{482}{\sqrt{1609}}
\end{split}
Best Answer
The line of shortest distance between the two given lines will be perpendicular to both. Therefore, you can take the cross product of the direction vectors to find the direction of shortest distance.
Now, you need to find two points on the two lines that are in that direction of each other. To do this, you could construct the plane including one of the given lines along with the direction vector of the segment of shortest distance. Then compute the intersection of that plane with the second line.