A variable plane passes through a fixed point(3,2,1) and meets x,y and z axes at A,B and C respectiv

Q: Why do we assume the plane equation is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$?

That is a great question! We use the intercept form because the problem explicitly defines the plane by its intersections with the $x, y,$ and $z$ axes at points $A, B,$ and $C$. This form is the most efficient way to relate the plane's position to the coordinate axes.

Q: Wait, why is the intersection point $(a, b, c)$?

Think about the equations of the planes we constructed: $x=a$ (parallel to $yz$-plane), $y=b$ (parallel to $zx$-plane), and $z=c$ (parallel to $xy$-plane). The intersection of these three planes is the unique point where all three conditions are met simultaneously, which is exactly $(a, b, c)$.

Q: Is there any restriction on $a, b, c$?

Yes, for the intercepts to exist, $a, b,$ and $c$ must be non-zero. This means the locus we find, $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$, naturally excludes the origin and the coordinate axes, which makes perfect physical sense!

Question

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 a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is :

Options:

(A) $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{11}{6}$

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

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

(D) $x + y + z = 6$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty 3D room with $x, y,$ and $z$ axes stretching to infinity. We consider a variable plane passing through a fixed point $P_0(3, 2, 1)$.

This plane is dynamic; it can tilt and rotate, provided it remains anchored to the point $(3, 2, 1)$. Our objective is to determine the locus of a point $P$ defined by the intercepts of this plane.

### The Intercept Form: The DNA of the Plane

When a plane intersects the $x, y,$ and $z$ axes at points $A, B,$ and $C$ respectively, we denote the distances from the origin as $a, b,$ and $c$. The equation of this plane is given by:

$$
\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1
$$

Since the plane is constrained to pass through the fixed point $(3, 2, 1)$, these coordinates must satisfy the plane's equation. Substituting these values, we obtain the fundamental constraint:

$$
\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1
$$

### Constructing the Invisible Box

We perform a geometric construction by drawing planes through the intercepts $A, B,$ and $C$ parallel to the coordinate planes. Specifically, we draw:

1. A plane $x = a$ parallel to the $yz$-plane.
2. A plane $y = b$ parallel to the $zx$-plane.
3. A plane $z = c$ parallel to the $xy$-plane.

These three planes intersect at a unique point $P(x, y, z)$. Consequently, the coordinates of this point are directly mapped to the intercepts such that $x = a$, $y = b$, and $z = c$.

### The Final Reveal

We now substitute our mapping $a = x$, $b = y$, and $c = z$ into our fundamental constraint equation. This substitution yields the locus of the point $P$:

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

This equation represents the path traced by the intersection point as the plane varies. The **locus of the point** is defined by the relation **$\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$**.

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Visualized Solution

The 3D Coordinate System & Fixed Point

Intercept Form of the Plane

Applying the Fixed Point Constraint

Coordinates of Intercepts

Constructing Parallel Planes

Equations of the New Planes

The Point of Intersection

Finding the Locus

Final Equation of the Locus

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Intercept Form

Constructing the Invisible Box

The Final Reveal

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