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A variable plane passes through a fixed point $(3, 2, 1)$ and meets x,y and z axes at A, B an C respectively. A plane is drawn parallel to yz-plane through A, a secound plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is

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a

$\frac{x}{3} + \frac{y}{2} + \frac{z}{1} = 1$

b

$x + y + z = 6$

c

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{11}{16}$

d

$\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$

answer is D.

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Detailed Solution

$\begin{array}{rcl} x & = & α, y = β z = γ \\ A (α, 0, 0) B (0, β, 0) C & = & (0, 0, γ) \\ \frac{x}{α} + \frac{y}{β} + \frac{z}{γ} & = & 1 \\ (x_{1}, y_{1}, z_{1}) & = & (α, β, γ) \\ \frac{3}{x_{1}} + \frac{2}{y_{1}} + \frac{z}{z_{1}} & = & 1 \end{array}$

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