The image of the line `(x-1)/3=(y-3)/1=(z-4)/(-5)` in the plane `2x-y+z+3=0` is the line (1) `(x+3)/3=(y-5)/1=(z-2)/(-5)` (2) `(x+3)/(-3)=(y-5)/(-1)=(z+2)/5` (3) `(x-3)/3=(y+5)/1=(z-2)/(-5)` (3) `(x-3)/(-3)=(y+5)/(-1)=(z-2)/5`

Find the image of the line (x-1)/3=(y-2)/4=(z-3)/5 on the plane x-2y+z+2=0 .

The distance between the line (x-2)/(1)=(y-3)/(1)=(z-4)/(5) and the plane 2x+3y-z+5=0 is

The angle between the line (x-2)/1=(y+3)/-2=(z+4)/-3 and the plane 2x-3y+z=5 is

The line (x-3)/1=(y-4)/2=(z-5)/2 cuts the plane x+y+z=17 at

The line (x-3)/1=(y-4)/2=(z-5)/2 cuts the plane x+y+z=17 at

Proved that the line (x-1)/2=(y+2)/(-3)=(z-3)/1 lies on the plane 7x+5y+z=0

The lines (x-2)/(1)=(y-3)/(2)=(z-4)/(3) and (x-1)/(-5)=(y-2)/(1)=(z-1)/(1) are

Perpendiculars are drawn from points on the line (x+2)/(2)=(y+1)/(-1)=(z)/(3) to the plane x+y+z=3 The feet of perpendiculars lie on the line (a) (x)/(5)=(y-1)/(8)=(z-2)/(-13) (b) (x)/(2)=(y-1)/(3)=(z-2)/(-5)(d)(x)/(4)=(y-1)/(-7)=(z-2)/(5)(d)(x)/(2)=(y-1)/(-7)=(z-2)/(5)