If the plane `2x-y+2z+3=0` has the distances `(1)/(3)` and `(2)/(3)` units from the planes `4x-2y+4z+lambda=0` and `2x-y+2z+mu=0`, respectively, then the maximum value of `lambda+mu` is equal to

Options

A. 13
B. 15
C. 5
D. 9

Answer

A

Explanation

Equation of given planes are
`2x-y+2z+3=0" "…(i)`
`4x-2y+4z+lambda=0" "...(ii)`
and `" "2x-y+2x+mu=0" "...(iii)`
`because`Distance between two parallel planes
`ax+by+cz+d_(1)=0`
and`" "ax+by+cz+d_(2)=0` is
distance`=(|d_(1)-d_(2)|)/(sqrt(a^(2)+b^(2)+c^(2)))`
`:.`Distance between planes (i) and (ii) is
`(|lambda-2(3)|)/(sqrt(16+4+16))=(1)/(3)" "["given"]`
`implies|lambda-6|=2implies" "lambda-6=+-2implieslambda=8" or "4`
and distance between planes (i) and (iii) is
`(|mu-3|)/(sqrt(4+1+4))=(2)/(3)" "["given"]`
`implies" "|mu-3|=2`
`implies" "mu-3=+-2impliesmu=5` or 1
So, maximum value of `(lambda+mu)` at `lambda=8` and `mu=5` it is equal to 13.