If the points $(1, 1, \lambda )$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0,$ then $\lambda$ satisfies the equation

$A$ straight line is given by $\vec{r} = (1 + t)\hat{i} + 3t\hat{j} + (1 - t)\hat{k}$ where $t \in R$. If this line lies in the plane $x + y + cz = d$,then the value of $(c + d)$ is

Let a unit vector $\hat{OP}$ make angles $\alpha, \beta, \gamma$ with the positive directions of the coordinate axes $OX, OY, OZ$ respectively,where $\beta \in (0, \frac{\pi}{2})$. If $\hat{OP}$ is perpendicular to the plane passing through the points $(1, 2, 3)$,$(2, 3, 4)$,and $(1, 5, 7)$,then which one of the following is true?

Let the points $P, Q$ and $R$ have position vectors $\overrightarrow{r_1} = 3i - 2j - k, \overrightarrow{r_2} = i + 3j + 4k$ and $\overrightarrow{r_3} = 2i + j - 2k$ respectively relative to an origin $O$. Then the distance of $P$ from the plane $OQR$ is:

The point $\bar{i}-2 \bar{j}$ lies on a line parallel to the vector $2 \bar{i}+\bar{k}$. The point $\bar{i}+2 \bar{j}$ lies on a plane parallel to the vectors $2 \bar{j}-\bar{k}$ and $\bar{i}+2 \bar{k}$. Find the point of intersection of the line and the plane.

Let the line $\ell: x = \frac{1-y}{-2} = \frac{z-3}{\lambda}, \lambda \in R$ meet the plane $P: x + 2y + 3z = 4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $\ell$ and the plane $P$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\alpha + 2\beta + 6\gamma$ is equal to