Let P be thte plane, which contains the line of intersection of the planws, `x + y + z -6 =0 and 2x + 3y + z + 5 =0` and it is perpendicular to the xy-plane. Then the distance of the point `(0,0,256)` from P is equal to `(m)/(sqrtn),` then the value of m,n is

Answer

To solve the problem step by step, we need to find the equation of the plane \( P \) that contains the line of intersection of the given planes and is perpendicular to the xy-plane. Then we will calculate the distance from the point \( (0, 0, 256) \) to this plane.

### Step 1: Identify the given planes
The two given planes are:
1. \( P_1: x + y + z - 6 = 0 \)
2. \( P_2: 2x + 3y + z + 5 = 0 \)

### Step 2: Write the family of planes
The equation of the family of planes containing the line of intersection of the two planes can be expressed as:
\[ P: P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes, we have:
\[ (x + y + z - 6) + \lambda(2x + 3y + z + 5) = 0 \]
This simplifies to:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y + (1 + \lambda)z + (5\lambda - 6) = 0 \]

### Step 3: Conditions for perpendicularity to the xy-plane
For the plane \( P \) to be perpendicular to the xy-plane, the coefficient of \( z \) must not be zero, and the coefficients of \( x \) and \( y \) must be zero. Thus, we set:
\[ 1 + 2\lambda = 0 \quad \text{and} \quad 1 + 3\lambda = 0 \]
From \( 1 + 2\lambda = 0 \):
\[ 2\lambda = -1 \implies \lambda = -\frac{1}{2} \]
From \( 1 + 3\lambda = 0 \):
\[ 3\lambda = -1 \implies \lambda = -\frac{1}{3} \]
Since both conditions cannot be satisfied simultaneously, we will only use the \( z \) coefficient condition.

### Step 4: Find \( \lambda \)
To ensure the plane is perpendicular to the xy-plane, we can choose a value of \( \lambda \). Let's set \( \lambda = -1 \) (as it simplifies our calculations):
\[ P: (1 + 2(-1))x + (1 + 3(-1))y + (1 - 1)z + (5(-1) - 6) = 0 \]
This simplifies to:
\[ -1x - 2y - 11 = 0 \implies x + 2y + 11 = 0 \]

### Step 5: Equation of the plane
The equation of the plane \( P \) is:
\[ x + 2y + 11 = 0 \]

### Step 6: Distance from the point to the plane
To find the distance \( d \) from the point \( (0, 0, 256) \) to the plane \( P \), we use the formula:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For the plane \( x + 2y + 11 = 0 \), we have:
- \( A = 1 \)
- \( B = 2 \)
- \( C = 0 \)
- \( D = 11 \)

Substituting the coordinates of the point \( (0, 0, 256) \):
\[ d = \frac{|1(0) + 2(0) + 0(256) + 11|}{\sqrt{1^2 + 2^2 + 0^2}} = \frac{|11|}{\sqrt{1 + 4}} = \frac{11}{\sqrt{5}} \]

### Step 7: Identify \( m \) and \( n \)
From the distance formula, we can see that:
\[ d = \frac{11}{\sqrt{5}} \]
Thus, \( m = 11 \) and \( n = 5 \).

### Final Answer
The values of \( m \) and \( n \) are:
- \( m = 11 \)
- \( n = 5 \)