सिद्ध कीजिए कि रेखा `(x-7)/(2)=(y+3)/(-1)=(z-4)/(1)` समतलों `6x+4y-5z=4` तथा `x-5y+2z=12` को एक उभयनिष्ठ बिन्दु पर काटती है।

Solve: x-y+2z=7 3x+4y-5z=-5 2x-y+3z=12

6x-9y-20z=-4 4x-15y+10z=-1 2x-3y-5z=-1

4x -5y-11z=12,. x-3y+z=1 2x + 3y-7z = 2

3x-4y+2z=-1 2x+3y+5z=7 x+z=2

3x-4y+2z=-1 2x+3y+5z=7 x+z=2

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) .