Statement-1: If V is the volume of a parallelopiped having three coterminous edges as $\vec a,\vec b$ and $\vec c$, then the volume of the parallelopiped having three coterminous edges as
$\vec α=(\vec a.\vec a) \vec a +(\vec a·\vec b) \vec b+(\vec a.\vec c) \vec c$
$\vec β=(\vec a.\vec b) \vec a + (\vec b.\vec b) \vec b + (\vec b.\vec c) \vec c$
$\vec γ=(\vec a.\vec c) \vec a + (\vec b.\vec c) \vec b + (\vec c.\vec c) \vec c$ is $V^3$.
Statement-2: For any three vectors $\vec a, \vec b, \vec c$
$\begin{vmatrix}\vec a.\vec a&\vec a.\vec b&\vec a.\vec c\\\vec b.\vec a&\vec b.\vec b&\vec b.\vec c\\\vec c.\vec a&\vec c.\vec b&\vec c.\vec c\end{vmatrix}=[\vec a\,\,\vec b\,\,\vec c]^3$
Answer & explanation
Correct answer: option 3
We have, $\left|[\vec a\,\,\vec b\,\,\vec c]\right|=V$
Let $V_1$ be the volume of the parallelopiped formed by the vectors $\vec α,\vec β$ and $\vec γ$. Then, $V_1 =\left|[\vec α\,\,\vec β\,\,\vec γ]\right|$
Now, $[\vec α\,\,\vec β\,\,\vec γ]=\begin{vmatrix}\vec a.\vec a&\vec a.\vec b&\vec a.\vec c\\\vec b.\vec a&\vec b.\vec b&\vec b.\vec c\\\vec c.\vec a&\vec c.\vec b&\vec c.\vec c\end{vmatrix}=[\vec a\,\,\vec b\,\,\vec c]$
$⇒[\vec α\,\,\vec β\,\,\vec γ]=[\vec a\,\,\vec b\,\,\vec c]^2[\vec a\,\,\vec b\,\,\vec c]$
$⇒[\vec α\,\,\vec β\,\,\vec γ]=[\vec a\,\,\vec b\,\,\vec c]^3$
$∴V_1=\left|[\vec α\,\,\vec β\,\,\vec γ]\right|=\left|[\vec a\,\,\vec b\,\,\vec c]^3\right|=V^3$
Statement-2 is not true.