Question

If the plane $2x - y + 2z + 3 = 0$ has the distances $\frac{1}{3} $ and $\frac{2}{3}$ 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:

• A. 15
• B. 5
• C. 13
• D. 9

Answer 1

Answer: (C)

$4x - 2y + 4z + 6 = 0 $
$ \frac{\left|\lambda-6\right|}{\sqrt{16+4+16}} = \left|\frac{\lambda-6}{6}\right| = \frac{1}{3}$
$ \left|\lambda-6\right| = 2 $
$ \lambda = 8,4 $
$ \frac{\left|\mu-3\right|}{\sqrt{4+4+1}} = \frac{2}{3}$
$ \left|\mu -3\right| = 2$
$ \mu=5, 1$
$ \therefore $ Maximum value of $ \left(\mu +\lambda\right) = 13 $