Statement-1: If $\vec a$ and $\vec b$ are non-collinear vectors, then points having position vectors $x_1\vec a +y_1\vec b, x_2\vec a + y_2\vec b$ and $x_3\vec a + y_3\vec b$ are collinear if $\begin{vmatrix}x_1& x_2 &x_3\\y_1 &y_2 &y_3\\1& 1&1\end{vmatrix}=0$

Statement-2: Three points with position vectors $\vec a,\vec b,\vec c$ are collinear iff there exist scalars $x, y, z$ not all zero such that $x\vec a+y\vec b +z\vec c = 0$, where $x + y + z = 0$.

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

Statement-2 is true.

Using statement-2, points $x_1\vec a +y_1\vec b, x_2\vec a + y_2\vec b$ and $x_3\vec a + y_3\vec b$ will be collinear iff there exist scalars $l, m, n$ such that

$l(x_1\vec a +y_1\vec b)+m(x_2\vec a + y_2\vec b)+n(x_3\vec a + y_3\vec b)=\vec{0}$, where $l+m+n=0$

$⇒(lx_1+mx_2 + nx_3) \vec{a} + (ly_1+ my_2+ny_3) \vec b=\vec 0$

$⇒lx_1 + mx_2 +nx_3 = \vec 0$ and $ly_1 + my_2 + ny_3=\vec 0$  [∵ $\vec a,\vec b$ where are non-collinear]

Thus, we have,

$lx_1 +mx_2 + nx_3 = 0$

$ly_1+ my_2+ny_3 = 0$

$l, m, n=0$

This is a homogeneous system of equations having non-trivial solutions (as $l, m, n$ are not all zero).

$∴\begin{vmatrix}x_1& x_2 &x_3\\y_1 &y_2 &y_3\\1& 1&1\end{vmatrix}=0$

So statement-1 is true and statement-2 is a correct explanation for statement-1.