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If the vectors $\vec{\alpha}=\hat{\mathbf{i}}+a \hat{\mathbf{j}}+\mathbf{a}^{2} \hat{\mathbf{k}}, \vec{\beta}=\hat{\mathbf{i}}+b \hat{\mathbf{j}}+\mathbf{b}^{2} \hat{\mathbf{k}}$, and $\vec{\gamma}=\hat{\mathbf{i}}+c \hat{\mathbf{j}}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}\mathbf{a} & \mathbf{a}^{2} & 1+\mathrm{a}^{3} \\ \mathbf{b} & \mathbf{b}^{2} & 1+\mathrm{b}^{3} \\ \mathrm{c} & \mathrm{c}^{2} & 1+\mathrm{c}^{3}\end{array}\right|=0$,
then the value of abc is
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then the value of abc is
Solution:
1418 Upvotes
Verified Answer
The correct answer is:
$-1$
Hint:
$\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^{2} & 1 \\ \mathrm{~b} & \mathrm{~b}^{2} & 1 \\ \mathrm{c} & \mathrm{c}^{2} & 1\end{array}\right|(1+\mathrm{abc})=0$
$a b c=-1[\because \vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar vector $]$
$\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^{2} & 1 \\ \mathrm{~b} & \mathrm{~b}^{2} & 1 \\ \mathrm{c} & \mathrm{c}^{2} & 1\end{array}\right|(1+\mathrm{abc})=0$
$a b c=-1[\because \vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar vector $]$
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