If three points A, B and C have position vect | vedclass

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(A) Given the position vectors of points $A, B, C$ as:$A = \hat{i} + x\hat{j} + 3\hat{k}$$B = 3\hat{i} + 4\hat{j} + 7\hat{k}$$C = y\hat{i} - 2\hat{j}

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$(2, -3)$

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(A) Given the position vectors of points $A, B, C$ as:$A = \hat{i} + x\hat{j} + 3\hat{k}$$B = 3\hat{i} + 4\hat{j} + 7\hat{k}$$C = y\hat{i} - 2\hat{j} - 5\hat{k}$Calculate the vectors $\vec{AB}$ and $\vec{BC}$:$\vec{AB} = B - A = (3-1)\hat{i} + (4-x)\hat{j} + (7-3)\hat{k} = 2\hat{i} + (4-x)\hat{j} + 4\hat{k}$$\vec{BC} = C - B = (y-3)\hat{i} + (-2-4)\hat{j} + (-5-7)\hat{k} = (y-3)\hat{i} - 6\hat{j} - 12\hat{k}$Since $A, B, C$ are collinear, $\vec{AB} = t\vec{BC}$ for some scalar $t$:$2\hat{i} + (4-x)\hat{j} + 4\hat{k} = t((y-3)\hat{i} - 6\hat{j} - 12\hat{k})$Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$:$1) \ 4 = -12t \Rightarrow t = -\frac{1}{3}$$2) \ 4 - x = -6t \Rightarrow 4 - x = -6(-\frac{1}{3}) = 2 \Rightarrow x = 2$$3) \ 2 = t(y-3) \Rightarrow 2 = -\frac{1}{3}(y-3) \Rightarrow -6 = y-3 \Rightarrow y = -3$Thus, $(x, y) = (2, -3)$.

If three points $A, B$ and $C$ have position vectors $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$ respectively and if they are collinear, then $(x, y)$ is

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