If the points A(1, 3, 5),B(2, 4, 6),and C(4, | vedclass

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(A) Given points are $A(1, 3, 5)$,$B(2, 4, 6)$,and $C(4, 5, k)$.First,calculate the squared distances between the points:$AB^2 = (2-1)^2 + (4-3)^2 + (

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$2$

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(A) Given points are $A(1, 3, 5)$,$B(2, 4, 6)$,and $C(4, 5, k)$.First,calculate the squared distances between the points:$AB^2 = (2-1)^2 + (4-3)^2 + (6-5)^2 = 1^2 + 1^2 + 1^2 = 3$$BC^2 = (4-2)^2 + (5-4)^2 + (k-6)^2 = 4 + 1 + (k-6)^2 = k^2 - 12k + 41$$AC^2 = (4-1)^2 + (5-3)^2 + (k-5)^2 = 9 + 4 + (k-5)^2 = k^2 - 10k + 38$Case $1$: Right-angled at $A$ $(AB^2 + AC^2 = BC^2)$$3 + k^2 - 10k + 38 = k^2 - 12k + 41$$41 - 10k = 41 - 12k$$2k = 0 \Rightarrow k = 0$Case $2$: Right-angled at $B$ $(AB^2 + BC^2 = AC^2)$$3 + k^2 - 12k + 41 = k^2 - 10k + 38$$44 - 12k = 38 - 10k$$6 = 2k \Rightarrow k = 3$Case $3$: Right-angled at $C$ $(AC^2 + BC^2 = AB^2)$$k^2 - 10k + 38 + k^2 - 12k + 41 = 3$$2k^2 - 22k + 76 = 0$$k^2 - 11k + 38 = 0$The discriminant $D = (-11)^2 - 4(1)(38) = 121 - 152 = -31 Thus,the possible values for $k$ are $0$ and $3$. The number of possible values is $2$.

If the points $A(1, 3, 5)$,$B(2, 4, 6)$,and $C(4, 5, k)$ form a right-angled triangle,then the number of possible values of $k$ is:

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