JEE Mains · Maths · STD 12 - 11. three dimension geometry
A variable plane passes through a fixed point \((3,2,1)\) and meets \(x, y\) and \(z\) axes at \(A, B\) and \(C\) respectively. A plane is drawn parallel to \(yz-\) plane through \(A\), a second 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
- A \((x +y+z = 6)\)
- B \(\frac{x}{3} + \frac{y}{2} + \frac{z}{1} = 1\)
- C \(\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1\)
- D \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{{11}}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1\)
Step-by-step Solution
Detailed explanation
If \(a, b, c\) are the intercepts of the variable plane on the \(x, y, z\) axes respectively, then the equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) And the point of intersection of the planes parallel to the \(x y, y z\) and \(z x\) planes is \((a, b,c)\)…
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