Prove that there are infinite planes containing both the lines $L_1,L_2$ and that there is a unique plane containing both the lines $L_3,L_4.$

Question

Let $L_1:\frac{x-1}{1}=\frac{y-0}{1}=\frac{z-2}{-5}$
$L_2:\frac{x-2}{2}=\frac{y-1}{2}=\frac{z+3}{-10}$
and let $L_3:\frac{x-0}{-6}=\frac{y-1}{9}=\frac{z-0}{-3}$
$L_4:\frac{x-1}{2}=\frac{y-4}{-3}=\frac{z-0}{1}$
Prove that there are infinite planes containing both the lines $L_1,L_2$ and that there is a unique plane containing both the lines $L_3,L_4.$

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As the direction ratios of $L_1$ and $L_2$ are proportional so $L_1$ and $L_2$ are parallel.So they are coplanar and lie in a plane.
Also the direction ratios of $L_3$ and $L_4$ are proportional so $L_3$ and $L_4$ are also parallel.So they are coplanar and lie in a plane.

But i do not know how to prove that there are infinite planes containing both the lines $L_1,L_2$ and that there is a unique plane containing both the lines $L_3,L_4.$

Please help me.

Answer 1

Nicholas has provided a good hint (but maybe there is a typo).

Another way for the first :

We have $$\frac{x-1}{1}=\frac{y-0}{1}\iff x-1=y\iff x-2=y-1\iff \frac{x-2}{2}=\frac{y-1}{2}$$ and $$\frac{y-0}{1}=\frac{z-2}{-5}\iff y-1=\frac{z+3}{-5}\iff \frac{y-1}{2}=\frac{z+3}{-10}$$ So, $L_1=L_2$.

For the second :

As you wrote, $L_3$ is parallel to $L_4$.

Note here that $L_3$ passes through $(0,1,0)$, but $L_4$ doesn't.

So $L_3\not=L_4$.