The distance from a point \( (1,1,1) \) to a variable plane \(\pi\) is 12 units and the points of intersections of the plane with X, Y, Z-axes are \( A, B, C \) respectively. If the point of intersection of the planes through the points \( A, B, C \) and parallel to the coordinate planes is \( P \), then the equation of the locus of \( P \) is:
The distance from a point \( (1,1,1) \) to a variable plane \(\pi\) is 12 units and the points of intersections of the plane with X, Y, Z-axes are \( A, B, C \) respectively. If the point of intersection of the planes through the points \( A, B, C \) and parallel to the coordinate planes is \( P \), then the equation of the locus of \( P \) is:
Show Hint
- \( \left( \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \right) = 143 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right) \)
- \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 144 \)
- \( \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)^2 = 144 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right) \)
- \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 144 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right)^2 \)
The Correct Option is C
Approach Solution - 1
Step 1: Equation of the Plane
The general equation of a plane passing through a given point \( (a, b, c) \) and having intercepts \( A, B, C \) on the X, Y, and Z axes respectively is: \[ \frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1. \] Given that the perpendicular distance from the point \( (1,1,1) \) to this plane is 12, we use the formula for the distance from a point to a plane: \[ \frac{|1/A + 1/B + 1/C - 1|}{\sqrt{(1/A)^2 + (1/B)^2 + (1/C)^2}} = 12. \] Squaring both sides and simplifying: \[ \left( \frac{1}{A} + \frac{1}{B} + \frac{1}{C} \right)^2 = 144 \left( \frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2} \right). \] Since \( P \) is the intersection of planes parallel to the coordinate planes passing through \( A, B, C \), its coordinates satisfy the same relation.
Final Answer: \( \boxed{\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)^2 = 144 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right)} \)
Approach Solution -2
To find the equation of the locus of \(P\), we need to consider the variable plane \(\pi\) which intersects the X, Y, and Z-axes at points \(A\), \(B\), and \(C\) respectively. The equation of the plane can be written as \( ax + by + cz = d \). The plane is at a distance of 12 units from the point \((1,1,1)\), and the formula for the distance \(D\) from a point \((x_1, y_1, z_1)\) to the plane \(ax + by + cz + d = 0\) is given by:
\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
Setting this distance to 12, we have:
\[ \frac{|a + b + c - d|}{\sqrt{a^2 + b^2 + c^2}} = 12 \]
From the coordinates \( x \)-intercept \( A \), \( y \)-intercept \( B \), and \( z \)-intercept \( C \), we derive \(d/a\), \(d/b\), and \(d/c\) respectively. Then, point \( P \), which is the intersection of planes parallel to coordinate planes passing through \( A, B,\) and \( C \) is at \((0,0,0)\). The coordinates of \( P \) can be expressed in terms of intercepts as \((d/a, d/b, d/c)\).
For the equation of the locus, given \( P = (d/a, d/b, d/c) \), we express it as:
\[ \frac{1}{x} = \frac{a}{d}, \frac{1}{y} = \frac{b}{d}, \frac{1}{z} = \frac{c}{d} \]
Using the known distance constraint, and combining expressions for intercepts, we get the equation:
\[ \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2 = 144\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) \]
This is the equation of the locus of \( P \).
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