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(D) Given position vectors are $\bar{a} = \hat{i} + 3\hat{j}$,$\bar{b} = 2\hat{i} + 5\hat{j}$,and $\bar{c} = 4\hat{i} + 2\hat{j}$.Given the relation $

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$-160$

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(D) Given position vectors are $\bar{a} = \hat{i} + 3\hat{j}$,$\bar{b} = 2\hat{i} + 5\hat{j}$,and $\bar{c} = 4\hat{i} + 2\hat{j}$.Given the relation $\bar{c} = t_{1}\bar{a} + t_{2}\bar{b}$,we substitute the vectors:$4\hat{i} + 2\hat{j} = t_{1}(\hat{i} + 3\hat{j}) + t_{2}(2\hat{i} + 5\hat{j})$$4\hat{i} + 2\hat{j} = (t_{1} + 2t_{2})\hat{i} + (3t_{1} + 5t_{2})\hat{j}$Comparing the coefficients of $\hat{i}$ and $\hat{j}$,we get the system of linear equations:$t_{1} + 2t_{2} = 4$  --- $(1)$$3t_{1} + 5t_{2} = 2$  --- $(2)$Multiply equation $(1)$ by $3$: $3t_{1} + 6t_{2} = 12$ --- $(3)$Subtract equation $(2)$ from $(3)$: $(3t_{1} + 6t_{2}) - (3t_{1} + 5t_{2}) = 12 - 2$$t_{2} = 10$Substitute $t_{2} = 10$ into equation $(1)$: $t_{1} + 2(10) = 4$$t_{1} + 20 = 4 \implies t_{1} = -16$Therefore,$t_{1}t_{2} = (-16)(10) = -160$.

If $\bar{a}, \bar{b}, \bar{c}$ are the position vectors of the points $A(1,3,0), B(2,5,0), C(4,2,0)$ respectively and $\bar{c}=t_{1} \bar{a}+t_{2} \bar{b}$,then the value of $t_{1} t_{2}$ is:

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