A variable plane passes through a fixed point(3,2,1) and meets and axes at A,B,C respectively. A plane is drawn parallel to plane through a second plane is drawn parallel to plane through A,B,C . Then the locus of the point of intersection of these three planes, is : (a) `1/x+1/y+1/z =11/6` (b) `x/3+y/2+z/1=1` (c) `3/x+2/y+1/z=1` (d) `xy+z=6`

Options

A. `1/x+1/y+1/z =11/6`
B. `x/3+y/2+z/1=1`
C. `3/x+2/y+1/z=1`
D. `xy+z=6`

Answer

To solve the problem step by step, we need to find the locus of the point of intersection of three planes: a variable plane passing through the fixed point (3, 2, 1) and the planes formed by the intersections with the coordinate axes.

### Step 1: Equation of the Variable Plane
Let the equation of the variable plane be given by:
\[ \frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1 \]
where A, B, and C are the x, y, and z-intercepts of the plane.

### Step 2: Condition for the Plane
Since the plane passes through the fixed point (3, 2, 1), we substitute these coordinates into the plane equation:
\[ \frac{3}{A} + \frac{2}{B} + \frac{1}{C} = 1 \]

### Step 3: Expressing A, B, and C
From the equation, we can express A, B, and C in terms of a parameter. Rearranging gives:
\[ \frac{3}{A} = 1 - \frac{2}{B} - \frac{1}{C} \]
This implies:
\[ A = \frac{3}{1 - \frac{2}{B} - \frac{1}{C}} \]

### Step 4: Finding the Planes through A, B, C
The points A, B, and C are the intercepts on the axes:
- A = (A, 0, 0)
- B = (0, B, 0)
- C = (0, 0, C)

The planes through these points that are parallel to the coordinate planes can be represented as:
1. Plane through A: \(x = A\)
2. Plane through B: \(y = B\)
3. Plane through C: \(z = C\)

### Step 5: Finding the Locus of Intersection
The locus of the intersection of these planes can be found by eliminating A, B, and C. We know:
\[ \frac{3}{A} + \frac{2}{B} + \frac{1}{C} = 1 \]
Rearranging gives:
\[ \frac{1}{A} = \frac{1}{3}(1 - \frac{2}{B} - \frac{1}{C}) \]

### Step 6: Substituting for A, B, and C
Substituting back into the equation gives us a relationship between x, y, and z:
\[ \frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1 \]

### Step 7: Final Form
Multiplying through by \(xyz\) gives:
\[ 3yz + 2xz + xy = xyz \]
Rearranging leads us to:
\[ xyz - 3yz - 2xz - xy = 0 \]

### Step 8: Identifying the Locus
This can be rearranged to find the locus:
\[ xy + z = 6 \]

### Conclusion
Thus, the locus of the point of intersection of the three planes is given by:
\[ \boxed{xy + z = 6} \]