A variable plane makes intercepts on X, Y and Z-axes and it makes a tetrahedron of volume 64cu. Units. The locus of foot of perpendicular from origin on this plane is

Options

A. (a)`(x^2+y^2+z^2)=384xyz`
B. (b)`xyz=681`
C. (c)`(x+y+z)((1)/(x)+(1)/(y)+(1)/(z))^2=16`
D. (d)`xyz(x+y+z)=81`

Answer

To solve the problem step by step, we need to find the locus of the foot of the perpendicular from the origin to a variable plane that makes intercepts on the X, Y, and Z axes and forms a tetrahedron of volume 64 cubic units.

### Step 1: Understand the equation of the plane
The equation of a plane in intercept form is given by:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
where \(a\), \(b\), and \(c\) are the intercepts on the X, Y, and Z axes, respectively.

### Step 2: Calculate the volume of the tetrahedron
The volume \(V\) of a tetrahedron formed by the coordinate axes and the plane is given by:
\[ V = \frac{1}{6} \cdot a \cdot b \cdot c \]
Given that the volume is 64 cubic units, we set up the equation:
\[ \frac{1}{6} \cdot a \cdot b \cdot c = 64 \]
Multiplying both sides by 6 gives:
\[ a \cdot b \cdot c = 384 \]

### Step 3: Find the locus of the foot of the perpendicular
Let \(P(x, y, z)\) be the foot of the perpendicular from the origin to the plane. The coordinates of \(P\) can be expressed in terms of \(k\) (a proportionality constant) as follows:
\[ x = \frac{k}{a}, \quad y = \frac{k}{b}, \quad z = \frac{k}{c} \]
From the equation of the plane, we can express \(k\) in terms of \(a\), \(b\), and \(c\):
\[ \frac{1}{k} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \]

### Step 4: Substitute and simplify
Substituting \(x\), \(y\), and \(z\) into the equation gives:
\[ \frac{1}{k} = \frac{b^2c^2 + a^2c^2 + a^2b^2}{(abc)^2} \]
Since \(abc = 384\), we can express \(k\) in terms of \(x\), \(y\), and \(z\):
\[ k^2 = \frac{(abc)^2}{b^2c^2 + a^2c^2 + a^2b^2} \]

### Step 5: Formulate the locus equation
Now, we can write:
\[ x^2 + y^2 + z^2 = k^2 \]
Substituting \(k^2\) into the equation gives:
\[ x^2 + y^2 + z^2 = \frac{(abc)^2}{b^2c^2 + a^2c^2 + a^2b^2} \]
Using \(a \cdot b \cdot c = 384\), we can express the locus as:
\[ x^2 + y^2 + z^2 = 384xyz \]

### Final Result
Thus, the locus of the foot of the perpendicular from the origin on the plane is given by:
\[ x^2 + y^2 + z^2 = 384xyz \]