The number of values of k for which the point | vedclass

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(C) Let the points be $A(-4, 9, k)$, $B(-1, 6, k)$, and $C(0, 7, 10)$.Calculate the squared distances between the points:$AB^2 = (-1 - (-4))^2 + (6 -

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$2$

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(C) Let the points be $A(-4, 9, k)$, $B(-1, 6, k)$, and $C(0, 7, 10)$.Calculate the squared distances between the points:$AB^2 = (-1 - (-4))^2 + (6 - 9)^2 + (k - k)^2 = 3^2 + (-3)^2 + 0^2 = 9 + 9 = 18$.$BC^2 = (0 - (-1))^2 + (7 - 6)^2 + (10 - k)^2 = 1^2 + 1^2 + (10 - k)^2 = 2 + (10 - k)^2$.$AC^2 = (0 - (-4))^2 + (7 - 9)^2 + (10 - k)^2 = 4^2 + (-2)^2 + (10 - k)^2 = 16 + 4 + (10 - k)^2 = 20 + (10 - k)^2$.For a right-angled isosceles triangle, two sides must be equal and the Pythagorean theorem must hold.Case $1$: $AB = BC$.$18 = 2 + (10 - k)^2 \implies (10 - k)^2 = 16 \implies 10 - k = \pm 4 \implies k = 6$ or $k = 14$.If $k = 6$, $AB^2 = 18$, $BC^2 = 18$, $AC^2 = 20 + 16 = 36$. Since $18 + 18 = 36$, $AB^2 + BC^2 = AC^2$, so it is a right-angled isosceles triangle.If $k = 14$, $AB^2 = 18$, $BC^2 = 18$, $AC^2 = 20 + 16 = 36$. Since $18 + 18 = 36$, it is a right-angled isosceles triangle.Case $2$: $AB = AC$.$18 = 20 + (10 - k)^2 \implies (10 - k)^2 = -2$, which is impossible.Case $3$: $BC = AC$.$2 + (10 - k)^2 = 20 + (10 - k)^2 \implies 2 = 20$, which is impossible.Thus, there are $2$ possible values for $k$.

The number of values of $k$ for which the points $A(-4, 9, k)$, $B(-1, 6, k)$, and $C(0, 7, 10)$ form a right-angled isosceles triangle is:

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