Question

The line $\frac{x-2}{4}=\frac{y+1}{-1}=\frac{z-1}{3}$ is perpendicular to the plane

$\left(a\right)2x-y+z=3$
$\left(b\right)4x-y+3z=1$
$\left(c\right)4x+y+3z=1$
$\left(d\right)x+7y+z=3$

Answer 1

A line and a plane is perpendicular if the direction ratio of the line is parallel to the normal vector of the plane.
$\frac{x-2}{4}=\frac{y+1}{-1}=\frac{z-1}{3}$
Direction ratios of the line are $(4,-1,3)$
The plane in option $\left(b\right)$ has the normal vector as $(4,-1,3)$

$\therefore$ Option $\left(b\right)$ is correct.