Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ and $\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$ be three non zero vectors such that $|\vec{c}|=1$, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ and $\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$, then $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=\lambda\left(a_1^2+a_2^2+a_3^2\right)\left(b_1^2+b_2^2+b_3^2\right)$ where $\lambda$ is equal to:

1/2

$\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|^2=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|^2$

$=|\vec{a} .(\vec{b} \times \vec{c})|^2$

$=|(\vec{a} \times \vec{b}) . \vec{c})\left.\right|^2$

$=(|\vec{a} \times \vec{b}||\vec{c}|)^2$

$=|\vec{a}|^2|\vec{b}|^2 \sin ^2 \frac{\pi}{4}$

$=\frac{1}{2}\left(a_1^2+a_2^2+a_3^2\right)\left(b_1^2+b_2^2+b_3^2\right)$

Hence (1) is correct answer.