Let `P_(1) : 2x + y + z + 1 = 0`
`P_(2) : 2x - y + z + 3 = 0` and `P_(3) : 2x + 3y + z + 5 = 0` be three planes, then the distance of the line of intersection of planes `P_(1) = 0` and `P_(2) = 0` from the plane `P_(3) = 0` is

Options

A. `3/(sqrt(14))` units
B. `6/(sqrt(14))` units
C. `3/(sqrt(7))` units
D. `6/(sqrt(7))` units

Answer

To find the distance of the line of intersection of the planes \( P_1 \) and \( P_2 \) from the plane \( P_3 \), we can follow these steps:

### Step 1: Write the equations of the planes

The equations of the planes are given as:
- \( P_1: 2x + y + z + 1 = 0 \)
- \( P_2: 2x - y + z + 3 = 0 \)
- \( P_3: 2x + 3y + z + 5 = 0 \)

### Step 2: Parameterize the line of intersection of \( P_1 \) and \( P_2 \)

To find the line of intersection of the planes \( P_1 \) and \( P_2 \), we can express one variable in terms of the others. Let's set \( x = t \) (a parameter).

Substituting \( x = t \) into the equations of \( P_1 \) and \( P_2 \):

1. From \( P_1 \):
\[ 2t + y + z + 1 = 0 \implies y + z = -2t - 1 \quad \text{(Equation 1)} \]

2. From \( P_2 \):
\[ 2t - y + z + 3 = 0 \implies -y + z = -2t - 3 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations

Now, we can solve Equations 1 and 2 simultaneously.

From Equation 1:
\[ z = -2t - 1 - y \]

Substituting this expression for \( z \) into Equation 2:
\[ -y + (-2t - 1 - y) + 3 = -2t - 3 \]
Simplifying:
\[ -2y - 2t + 2 = -2t - 3 \]
\[ -2y = -5 \implies y = \frac{5}{2} \]

Now substituting \( y = \frac{5}{2} \) back into Equation 1 to find \( z \):
\[ \frac{5}{2} + z = -2t - 1 \implies z = -2t - 1 - \frac{5}{2} = -2t - \frac{7}{2} \]

Thus, the parametric equations of the line of intersection are:
\[ x = t, \quad y = \frac{5}{2}, \quad z = -2t - \frac{7}{2} \]

### Step 4: Find a point on the line of intersection

We can choose \( t = 0 \) to find a specific point on the line:
\[ P(0, \frac{5}{2}, -\frac{7}{2}) = (0, \frac{5}{2}, -\frac{7}{2}) \]

### Step 5: Calculate the distance from point \( P \) to the plane \( P_3 \)

The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( ax + by + cz + d = 0 \) is given by:
\[ d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

For plane \( P_3: 2x + 3y + z + 5 = 0 \):
- \( a = 2, b = 3, c = 1, d = 5 \)
- Point \( P(0, \frac{5}{2}, -\frac{7}{2}) \)

Calculating the distance:
\[ d = \frac{|2(0) + 3\left(\frac{5}{2}\right) + 1\left(-\frac{7}{2}\right) + 5|}{\sqrt{2^2 + 3^2 + 1^2}} \]
\[ = \frac{|0 + \frac{15}{2} - \frac{7}{2} + 5|}{\sqrt{4 + 9 + 1}} \]
\[ = \frac{|0 + \frac{15 - 7 + 10}{2}|}{\sqrt{14}} = \frac{|8|}{\sqrt{14}} = \frac{8}{\sqrt{14}} = \frac{4\sqrt{14}}{7} \]

### Final Answer

Thus, the distance of the line of intersection of planes \( P_1 \) and \( P_2 \) from the plane \( P_3 \) is:
\[ \frac{8}{\sqrt{14}} \text{ units} \]