The points $A(a), B(b), C(c)$ will be collinear if

If $\vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{c}=-\hat{i}+\hat{j}-4 \hat{k}$ and $\vec{d}=\hat{i}+\hat{j}+\hat{k}$,then $|(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})|=$

Let $a, b$ and $c$ be unit vectors such that $a \cdot b = 0 = a \cdot c$ and the acute angle between $b$ and $c$ is $\frac{\pi}{3}$,then $|a \times b - a \times c|$ is equal to

If $a \neq 0, b \neq 0, c \neq 0, a \times b = 0$ and $b \times c = 0$,then $a \times c$ is equal to

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-7\hat{j}+\hat{k}$.

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.