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The problem asks to find the value of \( k \) such that triangle \( ABC \), with the given vertices \( A(1, -5, k) \), \( B(3, k, 1) \), and \( C(k, -5, 2) \), has a right angle at \( A \).

To solve this, we need to check the condition for a right angle at \( A \). This occurs when vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are perpendicular, i.e., their dot product is zero.

### Step 1: Find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).
- Vector \( \overrightarrow{AB} = B - A = (3 - 1, k - (-5), 1 - k) = (2, k + 5, 1 - k) \).
- Vector \( \overrightarrow{AC} = C - A = (k - 1, -5 - (-5), 2 - k) = (k - 1, 0, 2 - k) \).

### Step 2: Calculate the dot product \( \overrightarrow{AB} \cdot \overrightarrow{AC} \).
\[
\overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(k - 1) + (k + 5)(0) + (1 - k)(2 - k)
\]
This simplifies to:
\[
2(k - 1) + (1 - k)(2 - k) = 2k - 2 + (1 - k)(2 - k)
\]
Expanding \( (1 - k)(2 - k) \):
\[
(1 - k)(2 - k) = 2 - k - 2k + k^2 = k^2 - 3k + 2
\]
Now, substituting into the dot product equation:
\[
2k - 2 + k^2 - 3k + 2 = 0
\]
Simplifying:
\[
k^2 - k = 0
\]
Factoring the equation:
\[
k(k - 1) = 0
\]

### Step 3: Solve for \( k \).
From \( k(k - 1) = 0 \), we get two possible values for \( k \):
\[
k = 0 \quad \text{or} \quad k = 1
\]

Thus, the values of \( k \) that make the triangle \( ABC \) right-angled at \( A \) are \( k = 0 \) or \( k = 1 \).

Would you like further details on this solution or explanations of other methods?

Here are 5 related questions to expand on this:

1. What are the conditions for vectors to be perpendicular in 3D space?
2. How do you calculate the area of triangle \( ABC \) for \( k = 0 \) and \( k

Let A = (1, -5, k), B = (3, k, 1), and C = (k, -5, 2) be three vertices of a triangle ABC. Determine the value of k so that the triangle is right-angled at A.

In this problem, we determine the value of k that makes triangle ABC right-angled at vertex A. Using concepts of vectors and dot products, the solution involves finding when vectors AB and AC are perpendicular.

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