If $|a| = |b|$,then $(a + b) \cdot (a - b)$ is:

If the position vectors of the vertices $A, B, C$ of a triangle $ABC$ are $4 \hat{\imath} + 7 \hat{\jmath} + 8 \hat{k}$,$2 \hat{\imath} + 3 \hat{\jmath} + 4 \hat{k}$,and $2 \hat{\imath} + 5 \hat{\jmath} + 7 \hat{k}$ respectively,then the position vector of the point where the bisector of angle $A$ meets $BC$ is

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?

Let $ABC$ be a triangle and $P$ be a point inside $ABC$ such that $\overrightarrow{PA} + 2\overrightarrow{PB} + 3\overrightarrow{PC} = \vec{0}$. The ratio of the area of $\triangle ABC$ to that of $\triangle APC$ is

If $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}$,$\bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}$,$\bar{c}=\hat{i}+3 \hat{j}+\hat{k}$ and $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then $\lambda=$

If $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$,then $a$ is equal to