1.

The plane passing through the point (4, -1, 2) and perallel to the lines `(x+2)/(3)=(y-2)/(-1)=(z+1)/(2)` and `(x-2)/(1)=(y-3)/(2)=(z-4)/(3)` also passes through the pointA. (-1, -1, -1)B. (1, 1, -1)C. (1, 1, 1)D. (-1, -1, 1)

Answer» Correct Answer - C
Let a be the position vector of the given point (4,-1,2).
Then, `a=4hati-hatj+2hatk`
The direction vector of the lines
`(x+2)/(3)=(y-2)/(-1)=(z+1)/(2)` and `(x-2)/(1)=(y-3)/(2)=(z-4)/(3)`
are respectively
`b_(1)=3hati-hatj+2hatk`
and`" "b_(2)=hati+2hatj+3hatk`
Now, as the plane is parallel to both `b_(1)` and `b_(2)`
[`because` plane is parallel to the given lines]
So, normal vector (n) of the plane is perpendicular to both `b_(1)` and `b_(2)`.
`implies" "n=b_(1)xxb_(2)`and
Required equation of plane is
`(r-a).n=0`
`implies" "(r-a).(b_(1)xxb_(2))=0`
`implies|{:(x-4,y+1,z-2),(3,-1," "2),(1,2," "3):}|=0`
`[{:(becauser-a=(xhati+yhatj+zhatk)-(4hati-hatj+2hatk)),(=(x-4)hati+(y+1)hatj+(z-2)hatk),("and we know that"","[a b c]=a.(bxxc)),(|{:(a_(1),b_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|)]`
`implies(x-4)(-3-4)-(y+1)(9-2)+(z-2)(6+1)=0`
`implies" "-7(x-4)-7(y+1)+7(z-2)=0`
`implies" "(x-4)+(y+1)-(z-2)=0`
`implies" "x+y-z-1=0`
(1, 1, 1) is the only point that satisfies.


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