The plane denoted by `pi_(1) : 4x+7y+4z+81=0` is rotated through a right angle about its line of intersection with the plane `pi_(2) : 5x+3y+ 10 z = 25`. If the plane in its new position is denoted by `pi`, and the distance of this plane from the origin is `sqrtk` , where `k in N` , then k=

Explanation

To solve the problem step by step, we will follow the outlined procedure to find the value of \( k \).

### Step 1: Write the equations of the planes
We have two planes given:
1. Plane \( \pi_1: 4x + 7y + 4z + 81 = 0 \)
2. Plane \( \pi_2: 5x + 3y + 10z - 25 = 0 \)

### Step 2: Find the line of intersection of the two planes
The line of intersection of the two planes can be represented as:
\[ 4x + 7y + 4z + 81 + \lambda(5x + 3y + 10z - 25) = 0 \]
This simplifies to:
\[ (4 + 5\lambda)x + (7 + 3\lambda)y + (4 + 10\lambda)z + (81 - 25\lambda) = 0 \]
Let’s denote this equation as Equation (3).

### Step 3: Find the condition for perpendicularity
For the new plane (after rotation) to be perpendicular to the original plane \( \pi_1 \), the normal vectors must satisfy:
\[ (4 + 5\lambda, 7 + 3\lambda, 4 + 10\lambda) \cdot (4, 7, 4) = 0 \]
Calculating this gives:
\[ 4(4 + 5\lambda) + 7(7 + 3\lambda) + 4(4 + 10\lambda) = 0 \]
Expanding and simplifying:
\[ 16 + 20\lambda + 49 + 21\lambda + 16 + 40\lambda = 0 \]
Combining like terms:
\[ (20 + 21 + 40)\lambda + (16 + 49 + 16) = 0 \]
\[ 81\lambda + 81 = 0 \]
Thus, we find:
\[ \lambda = -1 \]

### Step 4: Substitute \( \lambda \) back into Equation (3)
Substituting \( \lambda = -1 \) into Equation (3):
\[ (4 - 5)x + (7 - 3)y + (4 - 10)z + (81 + 25) = 0 \]
This simplifies to:
\[ -x + 4y - 6z + 106 = 0 \]
Let’s denote this as Equation (4).

### Step 5: Find the distance of the new plane from the origin
The distance \( d \) from the origin to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our plane \( -x + 4y - 6z + 106 = 0 \):
- \( A = -1 \)
- \( B = 4 \)
- \( C = -6 \)
- \( D = 106 \)

Thus, the distance from the origin is:
\[ d = \frac{|106|}{\sqrt{(-1)^2 + 4^2 + (-6)^2}} = \frac{106}{\sqrt{1 + 16 + 36}} = \frac{106}{\sqrt{53}} \]

### Step 6: Express the distance in terms of \( \sqrt{k} \)
We have:
\[ d = \frac{106}{\sqrt{53}} = \sqrt{k} \]
Squaring both sides gives:
\[ \frac{106^2}{53} = k \]
Calculating \( 106^2 = 11236 \):
\[ k = \frac{11236}{53} = 212 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{212} \]