If the length of the projection of the line segment joining the points `(1, 2, -1)` and `(3, 5, 5)` on the plane `3x-4y+12z=5` is equal to d units, then the value of `169d^(2)` equal to

Answer

To solve the problem, we need to find the length of the projection of the line segment joining the points \( A(1, 2, -1) \) and \( B(3, 5, 5) \) onto the plane defined by the equation \( 3x - 4y + 12z = 5 \). We will denote the length of this projection as \( d \) and ultimately calculate \( 169d^2 \).

### Step-by-Step Solution:

1. **Find the Direction Vector of the Line Segment AB**:
The direction vector \( \vec{AB} \) can be found by subtracting the coordinates of point A from point B:
\[ \vec{AB} = B - A = (3 - 1, 5 - 2, 5 - (-1)) = (2, 3, 6) \]

2. **Identify the Normal Vector of the Plane**:
The normal vector \( \vec{n} \) of the plane \( 3x - 4y + 12z = 5 \) can be directly obtained from the coefficients of \( x, y, z \):
\[ \vec{n} = (3, -4, 12) \]

3. **Calculate the Magnitude of the Vectors**:
- Magnitude of \( \vec{AB} \):
\[ |\vec{AB}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
- Magnitude of \( \vec{n} \):
\[ |\vec{n}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \]

4. **Calculate the Dot Product of the Vectors**:
The dot product \( \vec{AB} \cdot \vec{n} \) is calculated as follows:
\[ \vec{AB} \cdot \vec{n} = (2)(3) + (3)(-4) + (6)(12) = 6 - 12 + 72 = 66 \]

5. **Calculate the Cosine of the Angle**:
Using the dot product to find \( \cos \theta \):
\[ \cos \theta = \frac{\vec{AB} \cdot \vec{n}}{|\vec{AB}| |\vec{n}|} = \frac{66}{7 \cdot 13} = \frac{66}{91} \]

6. **Calculate the Sine of the Angle**:
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{66}{91}\right)^2 = 1 - \frac{4356}{8281} = \frac{8281 - 4356}{8281} = \frac{3925}{8281} \]
Therefore, \( \sin \theta = \sqrt{\frac{3925}{8281}} \).

7. **Calculate the Length of the Projection**:
The length of the projection \( d \) is given by:
\[ d = |\vec{AB}| \sin \theta = 7 \cdot \sqrt{\frac{3925}{8281}} = \frac{7 \sqrt{3925}}{91} \]

8. **Calculate \( 169d^2 \)**:
First, find \( d^2 \):
\[ d^2 = \left(\frac{7 \sqrt{3925}}{91}\right)^2 = \frac{49 \cdot 3925}{8281} \]
Now, multiply by 169:
\[ 169d^2 = 169 \cdot \frac{49 \cdot 3925}{8281} = \frac{49 \cdot 169 \cdot 3925}{8281} = \frac{49 \cdot 3925}{1} = 192025 \]

### Final Answer:
Thus, the value of \( 169d^2 \) is \( 192025 \).