If the vectors $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$,then the value of $\lambda$ is

Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}$,$\vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c}+\hat{i}) \times (\vec{a}+\vec{b}+\hat{i}) = \vec{a} \times (\vec{c}+\hat{i})$ and $\vec{a} \cdot \vec{c} = -29$. Then $\vec{c} \cdot (-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

If $a = 3i - j + 2k$ and $b = 2i + j - k$,then evaluate $a \times (a \cdot b)$.

If $12 \hat{i}-12 \hat{j}-18 \hat{k}$,$-3 \hat{i}-6 \hat{j}-9 \hat{k}$ and $3 \hat{i}+3 \hat{j}-24 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$,then the position vector of the incentre of $\triangle ABC$ is

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$,then $|\vec{a} \times \vec{c}|$ is equal to

If $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$,$\bar{b}=\hat{i}-\hat{j}+\hat{k}$,and $\bar{c}=\hat{i}+\hat{j}-\hat{k}$,then a vector in the plane of $\bar{a}$ and $\bar{b}$,whose projection on $\bar{c}$ is $\frac{1}{\sqrt{3}}$,is