If the lines $\frac{1-x}{3} = \frac{y-2}{2\lambda} = \frac{z-3}{2}$ and $\frac{x-1}{3\lambda} = \frac{y-1}{1} = \frac{6-z}{5}$ are perpendicular, then $\lambda$ is equal to

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

If the points $P(3,\mu,6), Q(-1,-1,2)$ and $R(2,8,5)$ are collinear, then the value of '$\mu$' is

If the two lines given by

$x=py+q, z=ry+s$ and

$x=p'y+q', z=r'y+s'$

are perpendicular, then which of the following is equal to -1 ?

The direction ratios of the line $\frac{1-x}{3} = \frac{7y - 14}{2} = \frac{z-3}{2}$ are

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

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 :