Find the distance of a point `(2,4,-1)` from the line
`(x+5)/1=(y+3)/4=(z-6)/(-9)`.

Find the distance of the point (-2, 3, -5) from the line (x+2)/1=(y-3)/2=z/3.

Find the distance of the point (-2, 3, -5) from the line (x+2)/1=(y-3)/2=z/3.

The perpendicular distance of the point (2,4,-1) from line (x +5)/(1) =(y+3)/(4) = (z-6)/(-9) is

the distance of the point (2,3,4) from the line (1-x)=(y)/(2)=(1)/(3)(1+z)

the distance of the point (2,3,4) from the line (1-x)=y/2=1/3(1+z)

(i) Find the distance of the point (-2,3,-4) from the line : (x + 2)/(3) = (2y + 3)/(4) = (3z + 4)/(5), measured parallel to the plane 4x +12y -3z +1 = 0. (ii) Find the distance of the point -2hati+3hatj-4hatk from the line : vecr=hati+2hatj-hatk+lambda(hati+3hatj-9hatk) measured parallel to the plane x-y+2z-3=0

The distance of the point (-2,4,-5) from the line (x+3)/(3)=(y-4)/(5)=(z+8)/(6) is

The distance of the point (-2,4,-5) from the line (x+3)/(3)=(y-4)/(5)=(z+8)/(6) is