about to only mathematics

Options

A. `((7)/(3),(7)/(3),(5)/(3))`
B. `(-1, -1, 0)`
C. `(1, 1, 1)`
D. `((7)/(9),(7)/(9),(8)/(9))`

Answer

B, D

Explanation

Equation of straight line is `l`:`(x-x_(1))/(a)=(y-y_(1))/(b)=(z-z_(1))/( c )`
Since, `l` is perpendicular to `l_(1)` and `l_(2)`.
So, its DR's are cross-product of `l_(1)` and `l_(2)`.
Now, to find a point on `l_(2)` whose distance is given, assume a point and find its distance to obtain point.
Let`l:(x-0)/(alpha)=(y-0)/(b)=(z-0)/(c)`
which is perpendicular to
`l_(1):3hati-hatj+4hatk)divt(hati+2hatj+2hatk)`
`l_(2):(3hati+3hatj+2hatk)+s(2hati+2hatj+hatk)`
`:."DR's of "l" is"|{:(hati,hatj,hatk),(1,2,2),(2,2,1):}|=-2hati+3hatj-2hatk`
`l:(x)/(-2)=(y)/(3)=(z)/(-2)=k_(1),k_(2)`
Now, `A(-2k_(1),3k_(1),-2k_(1))" and "B(-2k_(2),3k_(2),-2k_(2)).`
Since, A lies on `l_(1)`.
`:." "(-2k_(1))hati+(3k_(1))hatj-(2k_(1))hatk=(3+t)hati+(-1+2t)hatj+(4+2t)hatk`
`implies" "3+t=-2k_(1),-1+2t=3k_(1),4+2t=-2k_(1)`
`:." "k_(1)=-1`
`implies" "A(2,-3,2)`
Let any point on `l_(2)(3+2s,3+2s,2+s)`
`sqrt((2-3-2s)^(2)+(-3-3-2s)^(2)+(2-2-s)^(2))=sqrt(17)`
`implies" "9s^(2)+28s+37=17`
`implies" "9s^(2)+28s+20=0`
`implies" "9s^(2)+18s+10s+20=0`
`implies" "(9s+10)(s+2)=0`
`:." "s=-2,(-10)/(9).`
Hence, `(-1,-1,0)"and "((7)/(9),(7)/(9),(8)/(9))` are required points.