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If the plane 2x-y+2x+3=0 has the distanc...

If the plane `2x-y+2x+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

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The correct Answer is:
7
`2x -y + 2z + 3 = 0` …(1)
`4x -2y + 4z + lamda = rArr 2x - y + 2z + (lamda)/(2) = 0`….(2)
Distance between (1) and (2)
`(|3 - (lamda)/(2)|)/(3) = (1)/(3) rArr |3 - (lamda)/(2)| =1`
`lamda = 8 or lamda = 4`
`2x - y + 2z + mu = 0` ....(3)
Distance between (1) and (3)
`(|3- mu|)/(3) = (2)/(3) rArr 3- mu =2 or 3 - mu = -2`
`mu =1 or mu = 5`
`(lamda + mu)_("max") = 13`
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