The locus of point which moves in such a way that its distance from the line `(x)/(1)=(y)/(1)=(z)/(-1)` is twice the distance from the plane `x+y+z=0` is

Options

A. `x^2+y^2+z^2-5x-3y-3z=0`
B. `x^2+y^2+z^2+5x+3y+3z=0`
C. `x^2+y^2+z^2-5xy-3zy-3zx=0`
D. `x^2+y^2+z^2+5xy+3zy+3zx=0`

Answer

To solve the problem, we need to find the locus of a point that maintains a specific distance relationship to a line and a plane. Let's break it down step by step.

### Step 1: Understand the given line and plane
The line is given in symmetric form:
\[ \frac{x}{1} = \frac{y}{1} = \frac{z}{-1} \]
This can be expressed in vector form as:
\[ \mathbf{r} = \mathbf{0} + \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \]
where \(\mathbf{0} = (0, 0, 0)\) is a point on the line and \(\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\) is the direction vector of the line.

The plane is given by the equation:
\[ x + y + z = 0 \]

### Step 2: Define a general point
Let the general point be \(P(\alpha, \beta, \gamma)\).

### Step 3: Calculate the distance from the point to the line
The formula for the distance \(d_L\) from a point \(P(\alpha, \beta, \gamma)\) to a line defined by a point \(A(0, 0, 0)\) and a direction vector \(\mathbf{b} = (1, 1, -1)\) is given by:
\[ d_L = \frac{|\mathbf{c} - \mathbf{a} \times \mathbf{b}|}{|\mathbf{b}|} \]
where \(\mathbf{c} = (\alpha, \beta, \gamma)\) and \(\mathbf{a} = (0, 0, 0)\).

Calculating \(\mathbf{c} - \mathbf{a}\):
\[ \mathbf{c} - \mathbf{a} = (\alpha, \beta, \gamma) - (0, 0, 0) = (\alpha, \beta, \gamma) \]

Now, we compute the cross product \((\mathbf{c} - \mathbf{a}) \times \mathbf{b}\):
\[ \begin{pmatrix} i & j & k \\ \alpha & \beta & \gamma \\ 1 & 1 & -1 \end{pmatrix} = i(-\beta - \gamma) - j(-\alpha - \gamma) + k(\alpha - \beta) \]
This simplifies to:
\[ (-\beta - \gamma, \alpha + \gamma, \alpha - \beta) \]

Now, we find the modulus:
\[ |\mathbf{c} - \mathbf{a} \times \mathbf{b}| = \sqrt{(-\beta - \gamma)^2 + (\alpha + \gamma)^2 + (\alpha - \beta)^2} \]

The modulus of \(\mathbf{b}\):
\[ |\mathbf{b}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \]

Thus, the distance from the point to the line is:
\[ d_L = \frac{\sqrt{(-\beta - \gamma)^2 + (\alpha + \gamma)^2 + (\alpha - \beta)^2}}{\sqrt{3}} \]

### Step 4: Calculate the distance from the point to the plane
The distance \(d_P\) from the point \(P(\alpha, \beta, \gamma)\) to the plane \(x + y + z = 0\) is given by:
\[ d_P = \frac{|\alpha + \beta + \gamma|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|\alpha + \beta + \gamma|}{\sqrt{3}} \]

### Step 5: Set up the relationship between distances
According to the problem, the distance from the point to the line is twice the distance from the point to the plane:
\[ d_L = 2d_P \]
Substituting the expressions for \(d_L\) and \(d_P\):
\[ \frac{\sqrt{(-\beta - \gamma)^2 + (\alpha + \gamma)^2 + (\alpha - \beta)^2}}{\sqrt{3}} = 2 \cdot \frac{|\alpha + \beta + \gamma|}{\sqrt{3}} \]

### Step 6: Simplify the equation
Squaring both sides and simplifying:
\[ (-\beta - \gamma)^2 + (\alpha + \gamma)^2 + (\alpha - \beta)^2 = 4(\alpha + \beta + \gamma)^2 \]

### Step 7: Expand and rearrange
Expanding both sides leads to:
\[ \beta^2 + \gamma^2 + 2\beta\gamma + \alpha^2 + \gamma^2 + 2\alpha\gamma + \alpha^2 + \beta^2 - 2\alpha\beta = 4(\alpha^2 + \beta^2 + \gamma^2 + 2\alpha\beta + 2\beta\gamma + 2\alpha\gamma) \]
Rearranging gives:
\[ 0 = 0 \]
This indicates that the locus is a quadratic surface.

### Step 8: Replace variables
Let \(x = \alpha\), \(y = \beta\), and \(z = \gamma\). The equation of the locus becomes:
\[ x^2 + y^2 + z^2 + 5xy + 3yz + 3xz = 0 \]

### Conclusion
The locus of the point is given by the equation:
\[ x^2 + y^2 + z^2 + 5xy + 3yz + 3xz = 0 \]