vectors
The normal of the plane is [-5,8,-14] and the direction vector of the line is [2,4,3]. I know that after using the equation cosθ=u⃗ ⋅v⃗ ||u⃗ ||⋅||v⃗ |, I must subtract 90º by the found angle in order to obtain the angle between the line and the plane.
The issue I'm having here is that the angle between the normal and the line is about 102.7º. I know the angle between the plane and the normal is exactly 90º, so how is this possible? How can I go about finding the plane-line angle?
The normal to the plane is ${\bf n}=(3,4,-1)$ as you have found. A vector in the direction of the line is ${\bf v}=(-2,3,-1)$. The angle between them is given by the dot product formula: $$\cos\theta=\frac{\bf n\cdot v}{|{\bf n}|\,|{\bf v}|}=\frac{7}{\sqrt{26}\sqrt{14}} =\frac{7}{2\sqrt{91}}=\frac{\sqrt{91}}{26}\ .$$ And the angle you want is $\frac\pi2-\theta$, draw a diagram and you will see why.
Honestly, between the direction vector $ v=(10,5,-5)$ and the normal vector $( 2,-1,4)$ the angle is not acute, in fact it is obtuse (you can see that by a simple plot). 
However as you are asking about the angle between a line and a plane, so the you must take care of the orientation of the vectors you are working with.
In you case, to find the angle $ \theta $ you can do the following : when finding $\cos x$, apply $\arccos$ to find an angle $\phi$. Then subtract $\phi $ from $180$ to get $ \alpha=180 - \phi$ . Now $\theta = 90 - \alpha$.
Best Answer
I have not done the computations, but I assume the issue is you can take two different vectors as the normal vector for the plane, and both works. Similarly you can take a different choice for the direction vector of the line. In practice one always choose the vector pairs such that the resulting angle is less or equal to 90 degrees. In your case the actual angle should be $$ 180-102.7=77.3 $$ and as a result the angle between the plane and the line should be $$ 90-77.3=12.7 $$ instead.