If $a=\hat{i}+\hat{j}+\hat{k}$, $c=\hat{j}-\hat{k}$, $a \times b=c$ and $a \cdot b=3$, then $b$ is equal to

Let a vector $\vec{a}$ have a magnitude $9$. Let a vector $\vec{b}$ be such that for every $(x, y) \in \mathbb{R} \times \mathbb{R} \setminus \{(0,0)\}$,the vector $(x \vec{a} + y \vec{b})$ is perpendicular to the vector $(6y \vec{a} - 18x \vec{b})$. Then the value of $|\vec{a} \times \vec{b}|$ is equal to:

If $|a|=1, |b|=2$ and the angle between $a$ and $b$ is $120^{\circ}$, then ${(a+3b) \times (3a-b)}^2$ is equal to

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

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

If $\vec{a}=\hat{i}-\hat{j}+\hat{k}$,$\vec{b}=\hat{i}+\hat{j}-2 \hat{k}$,$\vec{c}=2 \hat{i}-3 \hat{j}-\hat{k}$,and $\vec{d}=2 \hat{i}+\hat{j}+\hat{k}$ are four vectors,then find the value of $(\vec{a} \times \vec{c}) \times(\vec{b} \times \vec{d})$.