These Couple Lines

If $n=0$ you find quite quickly that all the direction cosines are $0$ and it is impossible to determine a line. So let's assume that $n\neq0$ .

Since the difference between direction numbers and direction cosines is a numerical factor, and both equation are homogeneous, we can use the given equations to find a triple of direction numbers, say ${L,M,N}$ with $N\neq 0$ .

We can arbitrarily choose $N=1$ and use the equations to find the other two direction numbers. We have

$3L+M+5=0\dots(1)$

$6M-2L+5LM=0\dots(2)$

It is not hard to use (1) to eliminate $M$ from (2) leading to

$L^2+3L+2=0$

and so $L=-2$ or $L=-1$

Using (1) we then see that the two possible trios of direction numbers are

${a}=(-2,1,1)$

${b}=(-1,-2,1)$

And so the cosine of the angle between the two lines is

$\dfrac{a.b}{\sqrt{a.a\times b.b}}=\dfrac{1}{6}$