The equation of the line passing through the points $(3, 2, 4)$ and $(4, 5, 2)$ is

  • A
    $\frac{x + 3}{1} = \frac{y + 2}{3} = \frac{z + 4}{-2}$
  • B
    $\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z - 4}{-2}$
  • C
    $\frac{x + 3}{7} = \frac{y + 2}{7} = \frac{z + 4}{6}$
  • D
    $\frac{x - 3}{7} = \frac{y - 2}{7} = \frac{z - 4}{6}$

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Find the foot of the perpendicular from the point $(2, 3, -8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also,find the perpendicular distance from the given point to the line.

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The angle between the two lines $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z+4}{2}$ and $\frac{x-5}{3}=\frac{y+2}{2}=\frac{z}{6}$ is $\qquad$ .

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}$ intersect each other,then $\lambda = \ldots$

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

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