$A$ unit vector perpendicular to the plane determined by the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ is:

Let $\overrightarrow{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$,$\overrightarrow{b} = 3 \hat{i} - \beta \hat{j} + 4 \hat{k}$ and $\overrightarrow{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $\alpha, \beta \in \mathbb{R}$,be three vectors. If the projection of $\overrightarrow{a}$ on $\overrightarrow{c}$ is $\frac{10}{3}$ and $\overrightarrow{b} \times \overrightarrow{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,then the value of $\alpha + \beta$ is equal to:

For some real number $\lambda$,if the area of the triangle having $\vec{a}=3 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{b}=\lambda \hat{i}+\hat{j}-3 \hat{k}$ as two of its sides is $\frac{\sqrt{195}}{2}$,then the number of distinct possible values of $\lambda$ is

The vector $x$ is perpendicular to the vectors $a=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $b=18 \hat{i}-22 \hat{j}-5 \hat{k}$,and it makes an obtuse angle with $\hat{j}$. If $|x|=14$,then $x=$

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k} .$ Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b},$ and $\vec{c} \cdot \vec{d}=15.$

Let $L_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}$ and $L_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $L_1$ and $L_2$ is