If overrightarrow{a} = 2hat{i} + hat{j} + 3ha | vedclass

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(A) Given $\overrightarrow{a} \cdot \overrightarrow{c} = 5 \Rightarrow 2c_{1} + c_{2} + 3c_{3} = 5$..........$(1)$Since $\overrightarrow{b} \perp \ove

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$150$

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(A) Given $\overrightarrow{a} \cdot \overrightarrow{c} = 5 \Rightarrow 2c_{1} + c_{2} + 3c_{3} = 5$..........$(1)$Since $\overrightarrow{b} \perp \overrightarrow{c}$,$\overrightarrow{b} \cdot \overrightarrow{c} = 0 \Rightarrow 3c_{1} + 3c_{2} + c_{3} = 0$.............$(2)$Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are coplanar,their scalar triple product is zero: $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$$\Rightarrow \begin{vmatrix} c_{1} & c_{2} & c_{3} \ 2 & 1 & 3 \ 3 & 3 & 1 \end{vmatrix} = 0$Expanding the determinant: $c_{1}(1 - 9) - c_{2}(2 - 9) + c_{3}(6 - 3) = 0$$\Rightarrow -8c_{1} + 7c_{2} + 3c_{3} = 0$ or $8c_{1} - 7c_{2} - 3c_{3} = 0$..............$(3)$Solving equations $(1), (2),$ and $(3)$:From $(2)$,$c_{3} = -3c_{1} - 3c_{2}$.Substitute into $(1)$: $2c_{1} + c_{2} + 3(-3c_{1} - 3c_{2}) = 5 \Rightarrow -7c_{1} - 8c_{2} = 5 \Rightarrow 7c_{1} + 8c_{2} = -5$.Substitute into $(3)$: $8c_{1} - 7c_{2} - 3(-3c_{1} - 3c_{2}) = 0 \Rightarrow 8c_{1} - 7c_{2} + 9c_{1} + 9c_{2} = 0 \Rightarrow 17c_{1} + 2c_{2} = 0 \Rightarrow c_{2} = -\frac{17}{2}c_{1}$.Substitute $c_{2}$ into $7c_{1} + 8c_{2} = -5$: $7c_{1} + 8(-\frac{17}{2}c_{1}) = -5 \Rightarrow 7c_{1} - 68c_{1} = -5 \Rightarrow -61c_{1} = -5 \Rightarrow c_{1} = \frac{5}{61} = \frac{10}{122}$.Then $c_{2} = -\frac{17}{2}(\frac{10}{122}) = -\frac{85}{122}$.Then $c_{3} = -3(\frac{10}{122}) - 3(-\frac{85}{122}) = \frac{-30 + 255}{122} = \frac{225}{122}$.Finally,$122(c_{1} + c_{2} + c_{3}) = 122(\frac{10 - 85 + 225}{122}) = 150$.

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