If $\vec{a} = \hat{i} - \hat{k}$, $\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$ and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ depends on

Let $\vec{v}=\alpha \hat{i}+2 \hat{j}-3 \hat{k}$,$\vec{w}=2 \alpha \hat{i}+\hat{j}-\hat{k}$,and $\vec{u}$ be a vector such that $|\vec{u}|=\alpha > 0$. If the minimum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is $-\alpha \sqrt{3401}$,and $|\vec{u} \cdot \hat{i}|^2=\frac{m}{n}$ where $m$ and $n$ are coprime natural numbers,then $m + n$ is equal to $.........$.

If the four points,whose position vectors are $3 \hat{i} - 4 \hat{j} + 2 \hat{k}$,$\hat{i} + 2 \hat{j} - \hat{k}$,$-2 \hat{i} - \hat{j} + 3 \hat{k}$,and $5 \hat{i} - 2 \alpha \hat{j} + 4 \hat{k}$ are coplanar,then $\alpha$ is equal to

If $\overline{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k})$ and $\overline{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k})$,then the value of $(2 \bar{a}-\bar{b}) \cdot [(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b})] = $

Unit vectors $a, b, c$ are coplanar. $A$ unit vector $d$ is perpendicular to the given vectors. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^{\circ}$,then $c = ....$

If the vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$,$2 \hat{i}+\hat{j}-\hat{k}$ and $\lambda \hat{i}-\hat{j}+2 \hat{k}$ are coplanar,then the value of $\lambda$ is: