If $\hat{a}, \hat{b}$ and $\hat{c}$ are non-coplanar vectors and if $\hat{d}$ is such that $\hat{d} = \frac{1}{x}(\hat{a} + \hat{b} + \hat{c})$ and $\hat{d} = \frac{1}{y}(\hat{b} + \hat{c} + \hat{d})$ where $x$ and $y$ are non-zero real numbers,then $\frac{1}{xy}(\hat{a} + \hat{b} + \hat{c} + \hat{d})$ equals to
- A$3\hat{c}$
- B$-\hat{a}$
- C$0$
- D$2\hat{a}$
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$A$. Unit vector in the direction opposite to that $a-b$ is $(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$ $B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$ $(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$ $C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is $(iii) \ -3 \hat{i} + 4 \hat{k}$ $D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$ $(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$ $(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$
| $A$. Unit vector in the direction opposite to that $a-b$ is | $(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$ |
| $B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$ | $(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$ |
| $C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is | $(iii) \ -3 \hat{i} + 4 \hat{k}$ |
| $D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$ | $(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$ |
| $(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$ |
If $\vec{a}+l \vec{b}+l^2 \vec{c}=0$ and $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=3(\vec{b} \times \vec{c})$, then the minimum value of such $l$ is
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