`P` is a point `(a, b, c).` Let `A, B, C` be images of Pin `y-z, z-x and x-y` planes respectively, then the equation of the plane `ABC` is

Options

A. Statement-I is true, Statement II is also true, Statement-II is the correct explanation of Statement-I.
B. Statement-I is true, Statement-II is also true, Statement-II is not the correct explanation of Statement-I.
C. Statement-I is true, Statement-II is false.
D. Statement-I is false, Statement -II is true.

Answer

To find the equation of the plane \( ABC \) formed by the images of the point \( P(a, b, c) \) in the \( y-z \), \( z-x \), and \( x-y \) planes, we will follow these steps:

### Step 1: Determine the coordinates of points A, B, and C
- **Point A** is the image of \( P \) in the \( y-z \) plane. In this plane, the \( x \)-coordinate is negated, so:
\[ A = (-a, b, c) \]

- **Point B** is the image of \( P \) in the \( z-x \) plane. In this plane, the \( y \)-coordinate is negated, so:
\[ B = (a, -b, c) \]

- **Point C** is the image of \( P \) in the \( x-y \) plane. In this plane, the \( z \)-coordinate is negated, so:
\[ C = (a, b, -c) \]

### Step 2: Find the direction ratios of vectors AB and AC
- The vector \( \overrightarrow{AB} \) can be calculated as:
\[ \overrightarrow{AB} = B - A = (a - (-a), -b - b, c - c) = (2a, -2b, 0) \]

- The vector \( \overrightarrow{AC} \) can be calculated as:
\[ \overrightarrow{AC} = C - A = (a - (-a), b - b, -c - c) = (2a, 0, -2c) \]

### Step 3: Use the normal vector to find the equation of the plane
- Let \( \vec{n} = (L, M, N) \) be the normal vector to the plane. The direction ratios of the normal vector can be found using the cross product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).

\[ \vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2a & -2b & 0 \\ 2a & 0 & -2c \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \left((-2b)(-2c) - (0)(0)\right) - \hat{j} \left((2a)(-2c) - (0)(2a)\right) + \hat{k} \left((2a)(0) - (-2b)(2a)\right) \]
\[ = \hat{i}(4bc) - \hat{j}(-4ac) + \hat{k}(4ab) \]
\[ = (4bc, 4ac, 4ab) \]

### Step 4: Write the equation of the plane
Using the point-normal form of the plane equation:
\[ L(x - x_0) + M(y - y_0) + N(z - z_0) = 0 \]
Substituting \( (x_0, y_0, z_0) = (-a, b, c) \) and \( (L, M, N) = (4bc, -4ac, 4ab) \):
\[ 4bc(x + a) - 4ac(y - b) + 4ab(z - c) = 0 \]
Simplifying:
\[ 4bcx + 4bca - 4acy + 4acb + 4abz - 4abc = 0 \]
\[ 4bcx - 4acy + 4abz + 4bca + 4acb - 4abc = 0 \]
Dividing through by 4:
\[ bcx - acy + abz + bca + acb - abc = 0 \]

### Final Equation of the Plane
The equation of the plane \( ABC \) is:
\[ bcx - acy + abz + bca + acb - abc = 0 \]