The equation of a plane containing the line $x-2=\frac{y-4}{4}=\frac{z-6}{7}$ and parallel to the line $\vec{r}=(\hat{i}+3\hat{j}+5\hat{k})+\lambda(3\hat{i}+5\hat{j}+7\hat{k})$ is

Let $L$ be the line of intersection of planes $\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=2$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=2$. If $P(\alpha, \beta, \gamma)$ is the foot of perpendicular on $L$ from the point $(1,2,0)$,then the value of $35(\alpha+\beta+\gamma)$ is equal to :

Find the length of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ and the coordinates of the foot of the perpendicular.

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$. Let $L_1: \overrightarrow{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \overrightarrow{a}, \lambda \in R$ and $L_2: \overrightarrow{r} = (\hat{j} + \hat{k}) + \mu \overrightarrow{b}, \mu \in R$ be two lines. If the line $L_3$ passes through the point of intersection of $L_1$ and $L_2$,and is parallel to $\overrightarrow{a} + \overrightarrow{b}$,then $L_3$ passes through the point:

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

The coordinates of the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane $2x - 3y - 6z = 49$ are