`(3/7z-1/2)/(z-1/4)-3/14=1/7`

Solve by matrix method 3/x+4/y+7/z=14 2/x-1/y+3/z=4 1/x+2/y-3/z=0

Find the S.D. between the lines : (i) (x)/(2) = (y)/(-3) = (z)/(1) and (x -2)/(3) = (y - 1)/(-5) = (z + 4)/(2) (ii) (x -1)/(2) = (y - 2)/(3) = (z - 3)/(2) and (x + 1)/(3) = (y - 1)/(2) = (z - 1)/(5) (iii) (x + 1)/(7) = (y + 1)/(-6) = (z + 1)/(1) and (x -3)/(1) = (y -5)/(-2) = (z - 7)/(1) (iv) (x - 3)/(3) = (y - 8)/(-1) = (z-3)/(1) and (x + 3)/(-3) = (y +7)/(2) = (z -6)/(4) .

The distance of point of intersection of lines (x-4)/1=(x+3)/-4=(z-1)/7 and (x-1)/2=(y+1)/-3=(z+10)/8 from (1,-4,7) is

The distance of point of intersection of lines (x-4)/(1)=(x+3)/(-4)=(z-1)/(7) and (x-1)/(2)=(y+1)/(-3)=(z+10)/(8) from (1,-4,7) is

Reflection of the line (x-1)/(-1)=(y-2)/(3)=(z-3)/(1) in the plane x+y+z=7 is

The lines (x-2)/(2)=(y)/(-2)=(z-7)/(16) and (x+3)/(4)=(y+2)/(3)=(z+2)/(1) intersect at the point "P" .If the distance of "P" from the line (x+1)/(2)=(y-1)/(3)=(z-1)/(1) is "l" ,then 14l^(2) is equal to