The equation of the plane, parallel to the plane $3x + 4y - 12z = 3$ and passes through $(1, 1, -1)$, is :

If the lines $x = ay+b$, $z = cy+d$ and $x = py+q$, $z = ry+s$ are perpendicular, then $ap + cr$ is equal to

Points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ at a distance of 5 units from the point $(1, 3, 3)$, are

Distance between the point (3, 4, 5) and the point where the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ meets the plane $x + y + z = 17$ is :

Let $(\alpha, 0, \beta)$ be the coordinates of the point, which divides the line segment joining the points (1,2,-3) and (1,-1,-4) internally, then $(\alpha^2+\beta^2)$ equals:

The shortest distance between the lines $\frac{x+3}{1} = \frac{y-2}{2} = \frac{z+4}{3}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4}$ is :

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