$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane defined by $\hat{i}$ and $\hat{i} + \hat{j}$,and the plane defined by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$. Find the angle between $\vec{a}$ and $\hat{i} - 2\hat{j} + 2\hat{k}$.

If a non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ and the plane determined by the vectors $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$,then the angle between the vectors $\vec{a}$ and $\hat{i}+\hat{j}+\hat{k}$ is

$a=3 \hat{i}+\hat{j}-\hat{k}, b=\hat{i}-4 \hat{j}+5 \hat{k}, c=4 \hat{i}+5 \hat{j}-\hat{k}$ are three vectors and a vector $r$ is perpendicular to both the vectors $b$ and $c$. If $r \cdot a=9$,then $r=$

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices of a triangle,show that $\frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]$ gives the vector area of the triangle. Hence,deduce the condition that the three points $\vec{a}, \vec{b}, \vec{c}$ are collinear. Also,find the unit vector normal to the plane of the triangle.

Let $2\hat{a} = \hat{b} \times \hat{c} + 2\hat{b}$. Then the sum of possible value$(s)$ of $\left| 2\hat{a} + \hat{b} + \hat{c} \right|$ is:

Let $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$,where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15(\alpha^{2} + 4)}$,then the value of $2|\vec{a}|^{2} + (\vec{a} \cdot \vec{b})|\vec{b}|^{2}$ is equal to