The planes `3x - y + z + 1 = 0 , 5x + y + 3z = 0 ` intersect in the line PQ. The equation of the plane through the point (2,1,4) and the perpendicular to PQ is

Options

A. `x +y - 2z = 5`
B. `x+y +2z = - 5`
C. `x + y +2z = 5`
D. `x + y - 2z =- 5`

Answer

To find the equation of the plane that passes through the point (2, 1, 4) and is perpendicular to the line of intersection PQ of the two given planes, we can follow these steps:

### Step 1: Identify the equations of the planes
The two planes are given as:
1. \( P_1: 3x - y + z + 1 = 0 \)
2. \( P_2: 5x + y + 3z = 0 \)

### Step 2: Find the normal vectors of the planes
The normal vector of plane \( P_1 \) is \( \vec{n_1} = (3, -1, 1) \) and the normal vector of plane \( P_2 \) is \( \vec{n_2} = (5, 1, 3) \).

### Step 3: Find the direction ratios of the line of intersection (PQ)
The direction ratios of the line of intersection can be found using the cross product of the normal vectors:
\[ \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 5 & 1 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{d} = \hat{i}((-1)(3) - (1)(1)) - \hat{j}((3)(3) - (1)(5)) + \hat{k}((3)(1) - (-1)(5)) \]
\[ = \hat{i}(-3 - 1) - \hat{j}(9 - 5) + \hat{k}(3 + 5) \]
\[ = \hat{i}(-4) - \hat{j}(4) + \hat{k}(8) \]
Thus, the direction ratios of line PQ are \( (-4, -4, 8) \) or simplified as \( (1, 1, -2) \).

### Step 4: Write the equation of the plane
The equation of a plane can be expressed as:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( (a, b, c) \) are the direction ratios of the normal to the plane.

Since we need a plane perpendicular to PQ, we can use the direction ratios \( (1, 1, -2) \) as the coefficients of \( x, y, z \) in the plane equation. The point through which the plane passes is \( (2, 1, 4) \).

Substituting these values into the plane equation gives:
\[ 1(x - 2) + 1(y - 1) - 2(z - 4) = 0 \]
Expanding this:
\[ x - 2 + y - 1 - 2z + 8 = 0 \]
Simplifying:
\[ x + y - 2z + 5 = 0 \]
or
\[ x + y - 2z = -5 \]

### Final Answer
The equation of the plane is:
\[ x + y - 2z = -5 \]