If A(2 hat{i} + hat{j} - hat{k}), B(lambda ha | vedclass

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(C) Given points are $A(2, 1, -1)$, $B(\lambda, 5, 4)$, $C(-4, 3, 2)$, and $D(-1, -2, 3)$.First, calculate the vectors:$\overrightarrow{AB} = (\lambda

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$7$

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(C) Given points are $A(2, 1, -1)$, $B(\lambda, 5, 4)$, $C(-4, 3, 2)$, and $D(-1, -2, 3)$.First, calculate the vectors:$\overrightarrow{AB} = (\lambda - 2) \hat{i} + (5 - 1) \hat{j} + (4 - (-1)) \hat{k} = (\lambda - 2) \hat{i} + 4 \hat{j} + 5 \hat{k}$$\overrightarrow{AC} = (-4 - 2) \hat{i} + (3 - 1) \hat{j} + (2 - (-1)) \hat{k} = -6 \hat{i} + 2 \hat{j} + 3 \hat{k}$$\overrightarrow{AD} = (-1 - 2) \hat{i} + (-2 - 1) \hat{j} + (3 - (-1)) \hat{k} = -3 \hat{i} - 3 \hat{j} + 4 \hat{k}$Given $\overrightarrow{AB} = x \overrightarrow{AC} + y \overrightarrow{AD}$, we equate the components:$(\lambda - 2) \hat{i} + 4 \hat{j} + 5 \hat{k} = x(-6 \hat{i} + 2 \hat{j} + 3 \hat{k}) + y(-3 \hat{i} - 3 \hat{j} + 4 \hat{k})$Comparing coefficients:$1) \lambda - 2 = -6x - 3y$$2) 4 = 2x - 3y$$3) 5 = 3x + 4y$Solving equations $(2)$ and $(3)$:From $(2)$, $3y = 2x - 4 \Rightarrow y = \frac{2x - 4}{3}$.Substitute into $(3)$: $5 = 3x + 4(\frac{2x - 4}{3}) \Rightarrow 15 = 9x + 8x - 16 \Rightarrow 17x = 31 \Rightarrow x = \frac{31}{17}$.Then $y = \frac{2(\frac{31}{17}) - 4}{3} = \frac{\frac{62 - 68}{17}}{3} = \frac{-6}{17 	imes 3} = -\frac{2}{17}$.Now substitute $x$ and $y$ into $(1)$:$\lambda - 2 = -6(\frac{31}{17}) - 3(-\frac{2}{17}) = \frac{-186 + 6}{17} = \frac{-180}{17}$$\lambda = 2 - \frac{180}{17} = \frac{34 - 180}{17} = -\frac{146}{17}$Finally, $17(\lambda + 9) = 17(-\frac{146}{17} + 9) = -146 + 153 = 7$.

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