1. Let the direction cosines of the required lines be $(l,m,n)$.

   They satisfy $l+3m+5n=0$ and $5lm-2mn+6nl=0.$

2. Since $l,m,n$ are direction cosines, only their ratios matter for determining the line. Let $x=\frac{l}{n},\qquad y=\frac{m}{n}$ (assuming $n\neq 0$; the solutions obtained will give valid directions).

   Then the first equation becomes $x+3y+5=0 \quad\Rightarrow\quad x=-3y-5.$

3. Substitute into the second equation: $5xy-2y+6x=0.$

   Using $x=-3y-5$, $5(-3y-5)y-2y+6(-3y-5)=0.$

   Simplify: $-15y^2-25y-2y-18y-30=0$ $-15y^2-45y-30=0.$

   Divide by $-15$: $y^2+3y+2=0$ $(y+1)(y+2)=0.$

   Hence, $y=-1 \quad \text{or} \quad y=-2.$

4. Find corresponding $x$ values:

   • If $y=-1$, then $x=-3(-1)-5= -2.$ So one direction ratio is $l:m:n = -2:-1:1.$

   • If $y=-2$, then $x=-3(-2)-5=1.$ So another direction ratio is $l:m:n = 1:-2:1.$

   Therefore the two lines have direction ratios $\vec d_1=(-2,-1,1), \qquad \vec d_2=(1,-2,1).$

5. Angle between the lines is the angle between these direction vectors.

   Compute dot product: $\vec d_1\cdot \vec d_2 = (-2)(1)+(-1)(-2)+(1)(1)=-2+2+1=1.$

   Magnitudes: $|\vec d_1|=\sqrt{(-2)^2+(-1)^2+1^2}=\sqrt{6},$ $|\vec d_2|=\sqrt{1^2+(-2)^2+1^2}=\sqrt{6}.$

   Hence

   =\frac{1}{\sqrt{6}\,\sqrt{6}}=\frac{1}{6}.$$ So, $$\theta=\cos^{-1}\left(\frac{1}{6}\right).$$

6. Checking options:

   • A: $\cos^{-1}(1/3)$ — incorrect
   • B: $\cos^{-1}(1/4)$ — incorrect
   • C: $\cos^{-1}(1/6)$ — correct
   • D: $\cos^{-1}(1/8)$ — incorrect

Therefore, the correct option is C.