$\bar{a} = \hat{i} + \hat{j} + \hat{k}$,$\bar{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\bar{c} = x\hat{i} + (x-1)\hat{j} - \hat{k}$. If the vector $\bar{c}$ lies in the plane of $\bar{a}$ and $\bar{b}$,then $x=$

If the volume of a parallelepiped whose coterminous edges are represented by the vectors $-12i + \alpha k$,$3j - k$,and $2i + j - 15k$ is $546$,find the value of $\alpha$.

$A$ unit vector coplanar with $i+j+3k$ and $i+3j+k$ and perpendicular to $i+j+k$ is

The volume of the tetrahedron formed by the coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is $3$. Then the volume of the parallelepiped formed by the coterminous edges $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ is

If $\hat{i}-3 \hat{j}+\hat{k}$ and $\lambda \hat{i}+3 \hat{j}$ are coplanar with a third vector,assuming the question implies the vectors are linearly dependent or part of a coplanar set,find $\lambda$. Given the standard form of such problems,if we consider the vectors $\vec{a} = \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \lambda \hat{i}+3 \hat{j}$ to be coplanar with a reference vector,let us assume the third vector is $\hat{k}$. For these to be coplanar,the scalar triple product must be zero: $\left|\begin{array}{ccc} 1 & -3 & 1 \\ \lambda & 3 & 0 \\ 0 & 0 & 1 \end{array}\right| = 0$. Solving this,$\lambda$ is equal to:

Let $O$ be the origin. Let $\overline{OP} = x\hat{i} + y\hat{j} - \hat{k}$ and $\overline{OQ} = -\hat{i} + 2\hat{j} + 3x\hat{k}$,where $x, y \in \mathbb{R}$ and $x > 0$,be such that $|\overline{PQ}| = \sqrt{20}$ and the vector $\overline{OP}$ is perpendicular to $\overline{OQ}$. If $\overline{OR} = 3\hat{i} + z\hat{j} - 7\hat{k}$,where $z \in \mathbb{R}$,is coplanar with $\overline{OP}$ and $\overline{OQ}$,then the value of $x^2 + y^2 + z^2$ is equal to ...... .