The straight line $\frac{x+3}{3} = \frac{y+2}{4} = \frac{z+1}{0}$ is

If the straight lines $x = 1 + s$, $y = -3 - \lambda s$, $z = 1 + \lambda s$ and $x = \frac{t}{2}$, $y = 1 + t$, $z = 2 - t$ with parameters $s$ and $t$ respectively, are coplanar, then $\lambda$ is equal to :

If two lines $\frac{x-3}{2} = \frac{y-4}{5} = \frac{z}{4}$ and $\frac{x-4}{3} = \frac{y-5}{6} = \frac{1-z}{k}$, are coplanar, then $k$ 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 :

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 value of $\lambda$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2\lambda} = \frac{z-3}{2}$ and $\frac{7-7x}{3\lambda} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angle, is:

Match List I with List II

Choose the correct answer from the options given below: