If the points P(4, 5, x),Q(3, y, 4) and R(5, | vedclass

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(C) The points $P(4, 5, x)$,$Q(3, y, 4)$,and $R(5, 8, 0)$ are collinear. Therefore,the vector $\vec{PQ}$ must be proportional to the vector $\vec{PR}$

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$4$

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(C) The points $P(4, 5, x)$,$Q(3, y, 4)$,and $R(5, 8, 0)$ are collinear. Therefore,the vector $\vec{PQ}$ must be proportional to the vector $\vec{PR}$. $\vec{PQ} = (3-4)\hat{i} + (y-5)\hat{j} + (4-x)\hat{k} = -\hat{i} + (y-5)\hat{j} + (4-x)\hat{k}$. $\vec{PR} = (5-4)\hat{i} + (8-5)\hat{j} + (0-x)\hat{k} = \hat{i} + 3\hat{j} - x\hat{k}$. Since the points are collinear,$\vec{PQ} = k \vec{PR}$ for some scalar $k$. $-\hat{i} + (y-5)\hat{j} + (4-x)\hat{k} = k(\hat{i} + 3\hat{j} - x\hat{k})$. Comparing the components: $1) -1 = k \Rightarrow k = -1$. $2) y - 5 = 3k \Rightarrow y - 5 = 3(-1) \Rightarrow y - 5 = -3 \Rightarrow y = 2$. $3) 4 - x = -kx \Rightarrow 4 - x = -(-1)x \Rightarrow 4 - x = x \Rightarrow 2x = 4 \Rightarrow x = 2$. Thus,$x + y = 2 + 2 = 4$.

If the points $P(4, 5, x)$,$Q(3, y, 4)$ and $R(5, 8, 0)$ are collinear,then the value of $x + y$ is

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