Statement-1: If the vectors $\vec{a}$ and $\vec{c}$ are non-collinear, then the lines $\vec{r} = 6 \vec{a} - \vec{c} + λ (2 \vec{c} - \vec{a})$ and $\vec{r} = \vec{a} - \vec{c} + \mu (\vec{a} + 3\vec{c})$ are coplanar.

Statement-2: There exist λ and $\mu $ scuh that the two values of $\vec{r}$ in statement-1 become same.

Statement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

We have,

$\vec{r} = 6\vec{a} - \vec{c} + λ (2\vec{c}-\vec{a})$ and $\vec{r} = \vec{a} - \vec{c} + \mu (\vec{a} + 3\vec{c})$

The two equations represent lines passing through points having position vectors $6\vec{a} - \vec{c}$ and $\vec{a} - \vec{c}$ respectively and parallel to vectors $2\vec{c} - \vec{a}$ and $\vec{a} + 3\vec{c}$ respectively. If these two lines intersect, then at the point of intersection $6\vec{a} - \vec{c}+λ(2\vec{c}- \vec{a}) = \vec{a} - \vec{c} + \mu (\vec{a} + 3\vec{c})$

$⇒ (5 - λ - \mu ) \vec{a} - (3 \mu - 2 λ) \vec{c} = \vec{0}$

$ ⇒ 5 - λ - \mu = 0 $ and $ 3 \mu - 2 λ = 0 $       $[∵ \vec{a}$ and $\vec{c}$ are non-collinear]

$ ⇒ λ = 3, \mu = 2 $

This means that for λ = 3 and $\mu = 2 $ the two values of $\vec{r}$ are same, So, statement-2 is true.

The truth of statement-2 also ensures that the two lines are intersecting hence coplanar.