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,lamda)` and equation of the plane be
`r*(3hati+2hatj+6hatk)=13`
`therefore(xhati+yhatj+zhatk)*(3hati+2hatj+6hatk)=13`
`[becauser*(xhati+yhatj+zhatk)]`
`implies3x+2y+6z=13`
`implies3x+2y+6z-13=0`
Now, distance of the planee from the point `(2,3,lamda)` is
`d|(ax_(1)+by_(1)+cz_(1)+d)/(sqrt(a^(2)+b^(2)+c^(2)))|`
`5=|(3xx2+2xx3+6xxlamda-13)/(sqrt(3^(2)+2^(2)+y^(2)))}` [given, d=5]
`+-5=(6+6+6lamda-13)/(sqrt(9+4+36))implies+-5=(6lamda-1)/(sqrt(49))`
`therefore5=|(6lamda-1)/(7)|`
`implies+-35=6lamda-1implies35=6lamda-1` or `-35=6lamda-1`
`implieslamda=6,lamda=-(17)/(3)`