Let the vectors vec{a}=(1+t) hat{i}+(1-t) hat | vedclass

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(D) Given that the vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0} \Rightarrow \alpha=\

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equal to $R - \{0\}$

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(D) Given that the vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0} \Rightarrow \alpha=\beta=\gamma=0$,it implies that the vectors are linearly independent.For three vectors to be linearly independent,their scalar triple product must be non-zero,i.e.,$[\vec{a} \vec{b} \vec{c}] 
eq 0$.Calculating the determinant:$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1+t & 1-t & 1 \ 1-t & 1+t & 2 \ 1 & -t & 1 \end{vmatrix}$Applying column operation $C_2 ightarrow C_1 + C_2$:$= \begin{vmatrix} 1+t & 2 & 1 \ 1-t & 2 & 2 \ 1 & 1-t & 1 \end{vmatrix}$Wait,let us re-evaluate the determinant directly:$[\vec{a} \vec{b} \vec{c}] = (1+t)((1+t)(1) - (2)(-t)) - (1-t)((1-t)(1) - (2)(1)) + 1((1-t)(-t) - (1+t)(1))$$= (1+t)(1+t+2t) - (1-t)(1-t-2) + (-t+t^2-1-t)$$= (1+t)(1+3t) - (1-t)(-1-t) + (t^2-2t-1)$$= (1+4t+3t^2) - (-(1-t)(1+t)) + t^2-2t-1$$= 1+4t+3t^2 + (1-t^2) + t^2-2t-1$$= 3t^2+2t+1$Re-checking the determinant calculation: $\begin{vmatrix} 1+t & 1-t & 1 \ 1-t & 1+t & 2 \ 1 & -t & 1 \end{vmatrix} = (1+t)(1+t+2t) - (1-t)(1-t-2) + 1(-t+t^2-1-t) = (1+t)(1+3t) - (1-t)(-1-t) + (t^2-2t-1) = 1+4t+3t^2 + (1-t^2) + t^2-2t-1 = 3t^2+2t+1$.Since the condition is $[\vec{a} \vec{b} \vec{c}] 
eq 0$,we have $3t^2+2t+1 
eq 0$. The discriminant $D = 2^2 - 4(3)(1) = 4 - 12 = -8 Thus,the set of all values of $t$ is $R$.

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