The coordinates of points A, B, C, D are (a, | vedclass

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(D) Given points are $A(a, 2, 1), B(1, -1, 1), C(2, -3, 4), D(a+1, a+2, a+3)$.First,calculate $AB = 5$:$AB = \sqrt{(a-1)^2 + (2 - (-1))^2 + (1-1)^2} =

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$-3$

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(D) Given points are $A(a, 2, 1), B(1, -1, 1), C(2, -3, 4), D(a+1, a+2, a+3)$.First,calculate $AB = 5$:$AB = \sqrt{(a-1)^2 + (2 - (-1))^2 + (1-1)^2} = 5$$\sqrt{(a-1)^2 + 3^2 + 0} = 5$$(a-1)^2 + 9 = 25$$(a-1)^2 = 16$$a-1 = \pm 4$$a = 5$ or $a = -3$  $(i)$Next,calculate $CD = 6$:$CD = \sqrt{(a+1-2)^2 + (a+2 - (-3))^2 + (a+3-4)^2} = 6$$\sqrt{(a-1)^2 + (a+5)^2 + (a-1)^2} = 6$$(a-1)^2 + (a+5)^2 + (a-1)^2 = 36$$(a^2 - 2a + 1) + (a^2 + 10a + 25) + (a^2 - 2a + 1) = 36$$3a^2 + 6a + 27 = 36$$3a^2 + 6a - 9 = 0$$a^2 + 2a - 3 = 0$$(a+3)(a-1) = 0$$a = -3$ or $a = 1$  $(ii)$The common solution for both $(i)$ and $(ii)$ is $a = -3$.

The coordinates of points $A, B, C, D$ are $(a, 2, 1), (1, -1, 1), (2, -3, 4)$ and $(a+1, a+2, a+3)$ respectively. If $AB = 5$ and $CD = 6$,then $a = $

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