If $a = 2 \hat{i} + 3 \hat{j} - \hat{k}$,$b = \hat{i} + 2 \hat{j} - 5 \hat{k}$,and $c = 3 \hat{i} + 5 \hat{j} - \hat{k}$,then a vector perpendicular to $a$ and in the plane containing $b$ and $c$ is:

If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0}$,then $\lambda$ and $\mu$ are respectively:

Let $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Assertion $(A)$ : The identity $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2=2|\overrightarrow{a}|^2$ holds for $\overrightarrow{a}$.
Reason $(R)$ : $\overrightarrow{a} \times \hat{i}=a_3 \hat{j}-a_2 \hat{k}$,$\overrightarrow{a} \times \hat{j}=a_1 \hat{k}-a_3 \hat{i}$,and $\overrightarrow{a} \times \hat{k}=a_2 \hat{i}-a_1 \hat{j}$.
Which of the following is correct?

Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \alpha \hat{k}$,where $\alpha > 0$. If the projection of $\vec{a} \times \vec{b}$ on the vector $\vec{c} = -\hat{i} + 2 \hat{j} - 2 \hat{k}$ is $30$,then $\alpha$ is equal to:

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

The two adjacent sides of a parallelogram are $2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\hat{i}-2 \hat{j}-3 \hat{k}.$ Find the unit vector parallel to its diagonal. Also,find its area.