let P be the plane, which contains the line of intersection of the planes `x+y+z-6=0` and `2x+3y+z+5=0` and it is perpendicular to the xy-plane thent he distance of the point (0,0,256) from P is equal to

Options

A. `17//sqrt(5)`
B. `63sqrt(5)`
C. `205sqrt(5)`
D. `11//sqrt(5)`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Identify the equations of the planes
We have two planes given by the equations:
1. Plane 1: \( P_1: x + y + z - 6 = 0 \)
2. Plane 2: \( P_2: 2x + 3y + z + 5 = 0 \)

### Step 2: Find the line of intersection of the two planes
The plane \( P \) that contains the line of intersection of these two planes can be expressed as a linear combination of the two planes:
\[ P: P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (x + y + z - 6) + \lambda(2x + 3y + z + 5) = 0 \]

### Step 3: Combine the equations
Expanding this, we get:
\[ x + y + z - 6 + 2\lambda x + 3\lambda y + \lambda z + 5\lambda = 0 \]
Combining like terms, we have:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y + (1 + \lambda)z + (-6 + 5\lambda) = 0 \]

### Step 4: Condition for the plane to be perpendicular to the xy-plane
For the plane \( P \) to be perpendicular to the xy-plane, the coefficient of \( z \) must be zero:
\[ 1 + \lambda = 0 \implies \lambda = -1 \]

### Step 5: Substitute \( \lambda \) back into the plane equation
Substituting \( \lambda = -1 \) into the equation of the plane:
\[ (1 - 2)x + (1 - 3)y + (1 - 1)z + (-6 - 5) = 0 \]
This simplifies to:
\[ -x - 2y - 11 = 0 \implies x + 2y + 11 = 0 \]

### Step 6: Identify the coefficients for distance calculation
The equation of the plane \( P \) can be rewritten as:
\[ x + 2y + 11 = 0 \]
Here, the coefficients are:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 0 \) (since there is no \( z \) term)
- \( k = 11 \)

### Step 7: Calculate the distance from the point \( (0, 0, 256) \)
The formula for the distance \( d \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + k = 0 \) is given by:
\[ d = \frac{|ax_1 + by_1 + cz_1 + k|}{\sqrt{a^2 + b^2 + c^2}} \]
Substituting \( (x_1, y_1, z_1) = (0, 0, 256) \):
\[ d = \frac{|1 \cdot 0 + 2 \cdot 0 + 0 \cdot 256 + 11|}{\sqrt{1^2 + 2^2 + 0^2}} = \frac{|11|}{\sqrt{1 + 4}} = \frac{11}{\sqrt{5}} \]

### Final Answer
The distance of the point \( (0, 0, 256) \) from the plane \( P \) is:
\[ \frac{11}{\sqrt{5}} \]