सरल रेखा `(x+3)/(3)=(y-2)/(-2)=(z+1)/(1)` व् समतल `4x+5y+3z-5=0` के प्रतिच्छेद बिन्दु ज्ञात कीजिए।

The line (x+3)/(3)=(y-2)/(-2)=(z+1)/(1) and the plane 4x+5y+3z-5=0 intersect at a point

The line (x+3)/(3)=(y-2)/(-2)=(z+1)/(1) and the plane 4x+5y+3z-5=0 intersect at a point

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) .

Show that the lines : (i) (x -5)/(7) = (y + 2)/(-5) = (z)/(1) " " and (x)/(1) = (y)/(2) = (z)/(3) (ii) (x - 3)/(2) = (y + 1)/(-3) = (z - 2)/(4) and (x + 2)/(2) = (y - 4)/(4) = (z + 5)/(2) are perpendicular to each other .

The lines (x-1)/(2)=(y)/(-1)=(z)/(2) and x-y+z-2=0=lambda x+3z+5 are coplanar for lambda

The (x+4)/(3) = (y+6)/(5) = (z-1)/(-2) and 3x - 2y + z+5 =0 = 2x + 3y + 4z -k are coplanar for k is equal to

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

A line (x-1)/(2)=(y-2)/(3)=(z-3)/(4) intersects the plane x-y+2z+2=0 at paint A .The equation of straight line passing through A Iying in the given plane and at minimum inclination with the given line is/are (A) (x+1)/(1)=(y+1)/(5)=(z+1)/(2) ,(B) 5x-y+4-0=2y-5z-3 ,(C) 5x+4-5z+1=0=2y-5z-3 ,(D) (x+2)/(1)=(y+6)/(5)=(z+3)/(2)