In $R^3$,consider the planes $P_1: y=0$ and $P_2: x+z=1$. Let $P_3$ be a plane,different from $P_1$ and $P_2$,which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0,1,0)$ from $P_3$ is $1$ and the distance of a point $(\alpha, \beta, \gamma)$ from $P_3$ is $2$,then which of the following relations is (are) true?
$(A)$ $2\alpha+\beta+2\gamma+2=0$
$(B)$ $2\alpha-\beta+2\gamma+4=0$
$(C)$ $2\alpha+\beta-2\gamma-10=0$
$(D)$ $2\alpha-\beta+2\gamma-8=0$

The angle between the line $\vec{r} = (\hat{i} + \hat{j} - 2\hat{k}) + \lambda (2\hat{i} - \hat{j} + \hat{k})$ and the normal to the plane $\vec{r} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 2$ is:

Let $P$ be the plane $\sqrt{3} x+2 y+3 z=16$ and let $S=\left\{\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}: \alpha^2+\beta^2+\gamma^2=1 \text{ and the distance of } (\alpha, \beta, \gamma) \text{ from the plane } P \text{ is } \frac{7}{2}\right\}$. Let $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$ be three distinct vectors in $S$ such that $|\overrightarrow{u}-\overrightarrow{v}|=|\overrightarrow{v}-\overrightarrow{w}|=|\overrightarrow{w}-\overrightarrow{u}|$. Let $V$ be the volume of the parallelepiped determined by vectors $\overrightarrow{u}, \overrightarrow{v}$ and $\overrightarrow{w}$. Then the value of $\frac{80}{\sqrt{3}} V$ is

The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

If the line,$\frac{x-3}{2}=\frac{y+2}{1}=\frac{z+4}{3}$ lies in the plane,$\ell x+m y-z=9$,then $\ell^2+m^2$ is equal to

The distance of the point $P(1, 2, 1)$ from the plane $2x + y - z = 10$ measured along the line $\frac{x - 5}{1} = \frac{2y - 3}{2} = \frac{z - \frac{5}{2}}{1}$ is