The equation of a line which is perpendicular to the plane 5x+2y−8z=0 and passing through -1,0,1 is

A
$\frac{x + 1}{5} = \frac{y}{2} = \frac{z - 1}{- 8}$
B
$\frac{x - 1}{5} = \frac{y - 1}{0} = \frac{z - 1}{- 8}$
C
$\frac{x + 1}{8} = \frac{z - 1}{2} = \frac{y - 0}{2}$
D
$\frac{x - 5}{- 1} = \frac{y - 2}{0} = \frac{z + 2}{1}$

Solution:

The equation of a line passing through the point $(x_{1}, y_{1}, z_{1})$ and perpendicular to the plane $a x + b y + c z + d = 0$ is $\frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$

Therefore, the equation of a line which is perpendicular to the plane $5 x + 2 y - 8 z = 0$ and passing through $(- 1, 0, 1)$ is $\frac{x + 1}{5} = \frac{y}{2} = \frac{z - 1}{- 8}$

The equation of a line which is perpendicular to the plane $5 x + 2 y - 8 z = 0$ and passing through $(- 1, 0, 1)$ is

Solution:

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