If the area of a parallelogram whose diagonals are represented by vectors $\vec{d_1} = 3 \hat{i} + \lambda \hat{j} + 2 \hat{k}$ and $\vec{d_2} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ is $\frac{\sqrt{117}}{2}$ sq. units,then $\lambda=$

In quadrilateral $ABCD$,$\overrightarrow{AB}=\vec{a}$,$\overrightarrow{BC}=\vec{b}$,$\overrightarrow{DA}=\vec{a}-\vec{b}$. $M$ is the midpoint of $BC$ and $X$ is a point on $DM$ such that $\overrightarrow{DX}=\frac{4}{5} \overrightarrow{DM}$. Then the points $A, X$ and $C$:

Let $\vec a = \hat i - \hat j,$ $\vec b = \hat i + \hat j + \hat k$ and $\vec c$ be a vector such that $\vec a \times \vec c + \vec b = 0$ and $\vec a \cdot \vec c = 4$,then ${\left| {\vec c} \right|^2}$ is equal to

Let $b = 4i + 3j$ and $c$ be two vectors perpendicular to each other in the $xy$-plane. All vectors in the same plane having projections $1$ and $2$ along $b$ and $c$ respectively,are given by

If the force $\overrightarrow{F} = \hat{i} + 2\hat{j} + 3\hat{k}$ moves a particle from position $\vec{r_1} = \hat{i} + \hat{j} - \hat{k}$ to $\vec{r_2} = 2\hat{i} - \hat{j} + \hat{k},$ then the work done is:

If the vectors $\hat{i}-2x\hat{j}-3y\hat{k}$ and $\hat{i}+3x\hat{j}+2y\hat{k}$ are orthogonal to each other,then the locus of the point $(x, y)$ is