The vlaue (s) of `lamda,` for the triangle with vertices `(6,10, 10),(1,0,-5) and (6,-10,lamda)` will be a right angled triangle is(are) :

Answer

To find the value(s) of \( \lambda \) for which the triangle with vertices \( A(6, 10, 10) \), \( B(1, 0, -5) \), and \( C(6, -10, \lambda) \) is a right-angled triangle, we will use the Pythagorean theorem. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

### Step-by-Step Solution:

1. **Calculate the distance \( AB \)**:
\[ AB = \sqrt{(6 - 1)^2 + (10 - 0)^2 + (10 - (-5))^2} \]
\[ = \sqrt{5^2 + 10^2 + 15^2} \]
\[ = \sqrt{25 + 100 + 225} = \sqrt{350} \]

2. **Calculate the distance \( BC \)**:
\[ BC = \sqrt{(1 - 6)^2 + (0 - (-10))^2 + (-5 - \lambda)^2} \]
\[ = \sqrt{(-5)^2 + (10)^2 + (-5 - \lambda)^2} \]
\[ = \sqrt{25 + 100 + (\lambda + 5)^2} \]
\[ = \sqrt{125 + (\lambda + 5)^2} \]

3. **Calculate the distance \( AC \)**:
\[ AC = \sqrt{(6 - 6)^2 + (10 - (-10))^2 + (10 - \lambda)^2} \]
\[ = \sqrt{0 + 20^2 + (10 - \lambda)^2} \]
\[ = \sqrt{400 + (10 - \lambda)^2} \]

4. **Apply the Pythagorean theorem**:
We can assume \( AC \) is the hypotenuse. Therefore, we have:
\[ AB^2 + BC^2 = AC^2 \]
Substituting the distances we calculated:
\[ 350 + (125 + (\lambda + 5)^2) = 400 + (10 - \lambda)^2 \]

5. **Expand and simplify**:
\[ 350 + 125 + (\lambda^2 + 10\lambda + 25) = 400 + (100 - 20\lambda + \lambda^2) \]
\[ 475 + \lambda^2 + 10\lambda + 25 = 500 - 20\lambda + \lambda^2 \]
The \( \lambda^2 \) terms cancel out:
\[ 475 + 10\lambda + 25 = 500 - 20\lambda \]
\[ 500 + 10\lambda = 500 - 20\lambda \]
\[ 10\lambda + 20\lambda = 0 \]
\[ 30\lambda = 0 \]
\[ \lambda = 0 \]

### Conclusion:
The value of \( \lambda \) for which the triangle is a right-angled triangle is \( \lambda = 0 \).