Statement-1: Let vec r be any vector in space. Then, vec r=(vec r.hat…

Since $\vec a, \vec b, \vec c$ are non-coplanar vectors.

Therefore, there exist scalars x, y, z such that

$\vec r = x\vec a+y\vec b +z\vec c$  ...(i)

Taking dot products with $\vec b×\vec c, \vec c×\vec a$ successively, we get

$\vec r.(\vec b×\vec c)=(x\vec a+y\vec b +z\vec c). (\vec b×\vec c)$

$\vec r.(\vec b×\vec c)=(x\vec a×y\vec b ×z\vec c). (\vec c×\vec a)$

$\vec r.(\vec b×\vec c)=(x\vec a+y\vec b +z\vec c). (\vec a×\vec b)$

$⇒[\vec r\,\,\vec b\,\,\vec c]=x[\vec a\,\,\vec b\,\,\vec c]$

$[\vec r\,\,\vec c\,\,\vec a]=y[\vec a\,\,\vec b\,\,\vec c]$

and, $[\vec r\,\,\vec a\,\,\vec b]=z[\vec a\,\,\vec b\,\,\vec c]$

$⇒x=\frac{[\vec r\,\,\vec b\,\,\vec c]}{[\vec a\,\,\vec b\,\,\vec c]},y=\frac{[\vec r\,\,\vec c\,\,\vec a]}{[\vec a\,\,\vec b\,\,\vec c]}$ and $z=\frac{[\vec r\,\,\vec a\,\,\vec b]}{[\vec a\,\,\vec b\,\,\vec c]}$

Substituting the values of $x, y, z$ in (i), we get

$\vec r=\left\{\frac{[\vec r\,\,\vec b\,\,\vec c]}{[\vec a\,\,\vec b\,\,\vec c]}\right\}\vec a+\left\{\frac{[\vec r\,\,\vec c\,\,\vec a]}{[\vec a\,\,\vec b\,\,\vec c]}\right\}\vec b+\left\{\frac{[\vec r\,\,\vec a\,\,\vec b]}{[\vec a\,\,\vec b\,\,\vec c]}\right\}\vec c$

So, statement-2 is true.

On replacing $\vec a, \vec b$ and $\vec c$ by $\hat i,\hat j$ and $\hat k$ respectively in statement-2, we obtain statement-1.

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