The angle between the line $\frac{x+2}{3} = \frac{y-3}{2} = \frac{z+5}{6}$ and the plane $2x + 10y - 11z = 5$ is:

If the lines $\frac{1-x}{2}=\frac{2y}{p}=\frac{z-1}{1}$ and $\frac{3-2x}{p}=\frac{y-1}{2}=\frac{z}{1}$ are parallel then value of 'p' is

The angle between the two planes $x + y - z = 3$ and $3x + 2y + z = 5$ is :

The equation of line of intersection of the planes $x + y + 3z = 7$ and $x - y + 2z = 3$ is:

The Cartesian equation of a line is $6x-2=3y+1=2z-2$. Then its vector from is:

If $\alpha, \beta$ and $\gamma$ are angle of inclinations of a line with x, y and z axes respectively, then the value of $2(\cos 2\alpha + \cos 2\beta + \cos 2\gamma)$ is

Let the equation of lines be as $L_1: \vec{r_1} = \vec{a_1} + \lambda\vec{b_1}$ and $L_2: \vec{r_2} = \vec{a_2} + \lambda\vec{b_2}$ such that $\vec{a_1} - \vec{a_2} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and $\vec{b_1} \times \vec{b_2} = 8\hat{i} - 4\hat{k}$. Then the shortest distance between $L_1$ and $L_2$ is