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Let `A(1,1,1),B(23,5)a n dC(-1,0,2)`be three points, then equation of a planeparallel to the plane `A B C`which is at distance isa. `2x-3y+z+2sqrt(14)=0`b. `2x-3y+z-sqrt(14)=0`c. `2x-3y+z+2=0`d. `2x-3y+z-2=0`A. `2x-3y+z+2sqrt14=0`B. `2x-3y+z-2sqrt14=0`C. `2x-3y+z+2=0`D. `2x-3y+z-2=0`

Answer» Correct Answer - a
A (1,1,1), B(2,3,5), C(-1,0,2) direction ratios of AB are `lt1,2,4gt`.
Direction ratios of AC are `lt-2,-1,1gt`. Therefore, direction ratios of normal to plane ABC are `lt-2,-3,1gt`
As a result, equation of the required plane be `2x-3y+z=0`.
Let the equation of the requried plane is `2x-3y+z=k`. Then
`|(k)/(sqrt(4+9+1))|=2ork=pm2sqrt14`
Hence, equation of the required plane is `2x-3y+z+2sqrt14=0`


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