If the vertices of a tetrahedron are $\vec{a} = \vec{j} + 2\vec{k}$,$\vec{b} = 3\vec{i} + \vec{k}$,$\vec{c} = 4\vec{i} + 3\vec{j} + 6\vec{k}$,and $\vec{d} = 2\vec{i} + 3\vec{j} + 2\vec{k}$,find its volume.

If $x \cdot a = 0, x \cdot b = 0$ and $x \cdot c = 0$ for some non-zero vector $x$,then the true statement is

If $a, b, c$ are any three non-coplanar vectors,then $[a + b, b + c, c + a] = $

If $\overline{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}, \overline{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$ and $\overline{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$ are non-zero non-coplanar vectors and $m$ is a non-zero scalar such that $[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = 28[\overline{a} \quad \overline{b} \quad \overline{c}]$,then the value of $m$ is:

Let the vectors $\overrightarrow{u}_1 = \hat{i} + \hat{j} + a\hat{k}$,$\overrightarrow{u}_2 = \hat{i} + b\hat{j} + \hat{k}$ and $\overrightarrow{u}_3 = c\hat{i} + \hat{j} + \hat{k}$ be coplanar. If the vectors $\overrightarrow{v}_1 = (a+b)\hat{i} + c\hat{j} + c\hat{k}$,$\overrightarrow{v}_2 = a\hat{i} + (b+c)\hat{j} + a\hat{k}$ and $\overrightarrow{v}_3 = b\hat{i} + b\hat{j} + (c+a)\hat{k}$ are also coplanar,then $6(a+b+c)$ is equal to $..............$.

Consider the vectors $\vec{x}=\hat{i}+2\hat{j}+3\hat{k}$,$\vec{y}=2\hat{i}+3\hat{j}+\hat{k}$,and $\vec{z}=3\hat{i}+\hat{j}+2\hat{k}$. For two distinct positive real numbers $\alpha$ and $\beta$,define $\vec{X}=\alpha\vec{x}+\beta\vec{y}-\vec{z}$,$\vec{Y}=\alpha\vec{y}+\beta\vec{z}-\vec{x}$,and $\vec{Z}=\alpha\vec{z}+\beta\vec{x}-\vec{y}$. If the vectors $\vec{X}, \vec{Y}$,and $\vec{Z}$ lie in a plane,the value of $\alpha+\beta-3$ is $....$.