Step 1: Define the direction vectors for the two lines.
Line 1 (x-axis) has direction ratios $(1,0,0)$. Let its direction vector be $\vec{u} = (1,0,0)$.
Line 2 has direction ratios $(3,-1,5)$. Let its direction vector be $\vec{v} = (3,-1,5)$.
Step 2: Find the unit vectors along these directions.
The direction of the angle bisectors is given by the sum and difference of the unit vectors of the original lines.
The magnitude of $\vec{u}$ is $|\vec{u}| = \sqrt{1^2+0^2+0^2} = 1$. The unit vector is $\hat{u} = (1,0,0)$.
The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{3^2+(-1)^2+5^2} = \sqrt{9+1+25} = \sqrt{35}$.
The unit vector is $\hat{v} = \frac{1}{\sqrt{35}}(3,-1,5)$.
Step 3: Find the direction vectors of the two angle bisectors.
The direction vectors of the bisectors are parallel to $\hat{u} + \hat{v}$ and $\hat{u} - \hat{v}$.
Bisector 1:
\[
\vec{d_1} \propto \hat{u} + \hat{v} = (1,0,0) + \frac{1}{\sqrt{35}}(3,-1,5) = \left(1+\frac{3}{\sqrt{35}}, -\frac{1}{\sqrt{35}}, \frac{5}{\sqrt{35}}\right).
\]
To get integer-like direction ratios, we can multiply by $\sqrt{35}$: $(\sqrt{35}+3, -1, 5)$. This corresponds to option (B).
Bisector 2:
\[
\vec{d_2} \propto \hat{u} - \hat{v} = (1,0,0) - \frac{1}{\sqrt{35}}(3,-1,5) = \left(1-\frac{3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, -\frac{5}{\sqrt{35}}\right).
\]
Multiplying by $\sqrt{35}$ to get simpler direction ratios: $(\sqrt{35}-3, 1, -5)$.
Step 4: Match the result with the corrected options.
The options in the image are highly corrupted. Option (C) appears to have the components $(\sqrt{35}-3, \dots, \dots)$. Our derived direction ratios are $(\sqrt{35}-3, 1, -5)$. Assuming typos in the option's second and third components, this matches the structure of the keyed answer (C).