https://d10lpgp6xz60nq.cloudfront.net/physics_images/PAT_CHE_0XI_B05_C11_E01_132_Q01.png

Answer

`a to q, s; b to r; c to p, q ; d to p`

Explanation

a. Line `" "(x-1)/(-2)= (y+2)/(3)= (z)/(-1) ` is along the vector `veca= -2hati+ 3hatj-hatk` and line `vecr= (3hati-hatj+hatk)+t(hati+hatj+hatk)` is along the vector `vecb = hati+hatj+hatk`. Here `veca bot vecb`.
Also `|{:(3-1,,-1-(-2),,1-0),(-2,,3,,-1),(1,,1,,1):}| ne 0`
b. The direction ratios of the line `x-y+ 2z-4=0`
`= 2x+y - 3z+5 =0` are `|{:(hati,,hatj,,hatk),(1,,-1,,2),(2,, 1,, -3) :}|= hati+7hatj+ 3hatk`.
Hence, the given two lines are parallel.
c. The given lines are `(x=t-3, y=-2t+1, z=-3t-2) and vecr= (t+1)hati+ (2t+3)hatj+ (-t-9)hatk`, or
`(x+3)/(1)= (y-1)/(-2) = (z+2)/(-3) and (x-1)/(1)= (y-3)/(2)= (z+9)/(-1)`.
The lines are perpendicular as
`(1)(1)+(-2)(2)+ (-3)(-1)=0`.
Also `|{:(-3-1,,1-3,,-2-(-9)),(1,,-2,,-3),(1,,2,,-1):}|=0`
Hence, the lines are intersecting.
d. The given lines are `vecr= (hati+3hatj-hatk)+t(2hati-hatj-hatk) and vecr= (-hati-2hatj+5hatk)+s(hati-2hatj+ (3)/(4) hatk)`.
`|{:(1-(-1),,3-(-2),,-1-5),(2,,-1,,-1),(1,,-2,,3//4):}|=0`
Hence, the lines are coplanar and intersecting (as the lines are not parallel).