ABC is a triangle right-angled at B . If A(k, 1, -1) , B(2k, 0, 2)...

The direction ratios of the line segment $AB$ are $(k - 2k, 1 - 0, -1 - 2) = (-k, 1, -3)$.

The direction ratios of the line segment $BC$ are $(2+2k - 2k, k - 0, 1 - 2) = (2, k, -1)$.

Since the triangle is right-angled at $B$, the line segments $AB$ and $BC$ are perpendicular to each other.

Therefore, the dot product of their direction ratios is zero.

$(-k)(2) + (1)(k) + (-3)(-1) = 0$

$-2k + k + 3 = 0$

$k = 3$

Answer: $3$