$ABC$ is a triangle right-angled at B. If A(k,1,−1), B(2k,0,2) and $C(2+2k,k,1)$ are the vertices of the triangle, then what is the value of k?

Calculation:

Given,

Points A(k, 1, -1), B(2k, 0, 2), and C(2 + 2k, k, 1) are the vertices of the triangle.

The vectors AB and BC are given by:

\( \overrightarrow{AB} = B - A = (2k - k, 0 - 1, 2 - (-1)) = (k, -1, 3) \)

\( \overrightarrow{BC} = C - B = (2 + 2k - 2k, k - 0, 1 - 2) = (2, k, -1) \)

The dot product of \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) is:

\( \overrightarrow{AB} \cdot \overrightarrow{BC} = (k)(2) + (-1)(k) + (3)(-1) = 2k - k - 3 = k - 3 \)

Since the vectors are perpendicular, their dot product must be zero:

\( k - 3 = 0 \)

Thus, \( k = 3 \).

∴ The value of k is \(3\).

Hence, the correct answer is Option 4.