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 $....$.

If $p = \frac{b \times c}{[a, b, c]}, q = \frac{c \times a}{[a, b, c]}, r = \frac{a \times b}{[a, b, c]}$,where $a, b, c$ are three non-coplanar vectors,then the value of $(a + b + c) \cdot (p + q + r)$ is given by

If the position vectors of the vertices $A, B$ and $C$ are $6i$,$6j$ and $k$ respectively with respect to the origin $O$,then the volume of the tetrahedron $OABC$ is

If $\bar{a}=\hat{i}+\hat{j}+\hat{k}$,$\bar{b}=4\hat{i}+3\hat{j}+4\hat{k}$,and $\bar{c}=\hat{i}+\alpha\hat{j}+\beta\hat{k}$ are linearly dependent vectors and $|\bar{c}|=\sqrt{3}$,then the values of $\alpha$ and $\beta$ are respectively.

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector, let us assume the vectors are $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$, $\vec{b} = \lambda \hat{i}+3 \hat{j}$, and we consider the standard basis vectors or a third vector to define coplanarity. However, if the question implies these two vectors are coplanar with the origin or a specific plane, we evaluate the scalar triple product. Given the standard interpretation of such problems, if $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ are coplanar with $\vec{c} = \hat{j}$, then the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$. Solving for $\lambda$ where $\vec{a} = (1, -3, 1)$, $\vec{b} = (\lambda, 3, 0)$, and $\vec{c} = (0, 1, 0)$:

If $a=x \hat{i}+y \hat{j}+z \hat{k}$, then $(a \times \hat{i}) \cdot(\hat{i}+\hat{j})+(a \times \hat{j}) \cdot(\hat{j}+\hat{k})+(a \times \hat{k}) \cdot(\hat{k}+\hat{i})=$