If the distance of points `2hati+3hattj+lamdahatk` from the plane `r*(3hati+2hatj+6hatk)=13` is 5 units, then `lamda=`

Options

A. `6,-(17)/(3)`
B. `6,(17)/(3)`
C. `-6,-(17)/(3)`
D. `-6,(17)/(3)`

Answer

a

Explanation

Given point be `(2,3,lambda)` and equation of the plane be
`r. (3hati +2hatj+6hatk) = 13`
`:. (xhati + yhatj + zhatk).(3hati + 2hatj + 6hatk) = 13`
`[ :' r . (xhati + yhatj + zhatk)]`
`rArr 3x+ 2y +6z = 13`
or `3x+ 2 y + 6z - 13 = 0`
Now, distance of the plane from the point `(2,3, lambda)` is
`d=|(ax_(1)+by_(1)+cz_(1)+d)/(sqrt(a^(2)+b^(2)+c^(2)))|`
`5=|(3xx2+2xx3+6xxlambda - 13)/(sqrt(3^(2)+2^(2)+6^(2)))|`
[given ,`d = 5`]
`+- 5 = (6+ 6 +6lambda - 13)/(sqrt(9+4+36))`
`rArr +-5=(6lambda -1)/(sqrt(49))`
`:. 5 = |(6lambda - 1)/(7)|`
`rArr +-35=6lambda - 1`
`rArr 35=6lambda - 1`
or `-35=6lambda - 1`
`rArr lambda = 6, lambda = - 17/3`