If $\vec{a} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k}$,then $(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = \dots$

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices $A, B, C$ of a triangle respectively. Find the area of the triangle.

$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is

$A$ particle acted on by constant forces $4i + j - 3k$ and $3i + j - k$ is displaced from the point $i + 2j + 3k$ to the point $5i + 4j + k$. The total work done by the force is ............... $unit$.

If $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - (\lambda^2 + 3\lambda)\hat{k}$ (where $\lambda$ is a constant) and $\vec{a}$ is perpendicular to $\vec{c} - \lambda\vec{b}$,then the sum of different values of $\lambda$ is:

Two forces $\vec{F_1} = 2\hat{i} - 5\hat{j} + 6\hat{k}$ and $\vec{F_2} = -\hat{i} + 2\hat{j} - \hat{k}$ act on a particle. The particle is displaced from point $P(4\hat{i} - 3\hat{j} + 2\hat{k})$ to point $Q(6\hat{i} + \hat{j} + 3\hat{k})$. The work done by the forces is ............. units.