Let the vectors a=(1+t) i+(1-t) j+ k, b=(1-t) i+(1+t) j+2 k and =t i-t j+ k,t∈ R be such that for α ,β ,γ ∈ R,α a+β b+γ = 0 ⇒ α =β =γ =0. Then, the…

Given the vectors $\vec{a}=(1+t)\hat{i}+(1-t)\hat{j}+\hat{k}$a=(1+t)i^+(1−t)j^​+k^, $\vec{b}=(1-t)\hat{i}+(1+t)\hat{j}+2\hat{k}$b=(1−t)i^+(1+t)j^​+2k^ and $\vec{c}=t\hat{i}-t\hat{j}+\hat{k},t\in R$c=ti^−tj^​+k^,t∈R be such that for $\alpha ,\beta ,\gamma \in R,\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\vec{0}$α,β,γ∈R,αa+βb+γc=0 $\Rightarrow \alpha =\beta =\gamma =0$⇒α=β=γ=0. Then, the set of all values of $t$t is

By its given condition $:\vec{a},\vec{b},\vec{c}$:a,b,care linearly independent vectors or they are non-coplanar

We know that when vectors are non-coplanar then, $[\begin{matrix}\vec{a} & \vec{b} & \vec{c}\end{matrix}]\neq 0...(1)$[a​b​c​]=0...(1)

Now, $[\begin{matrix}\vec{a} & \vec{b} & \vec{c}\end{matrix}]$[a​b​c​]

$$=|\begin{matrix}1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1\end{matrix}|$$=∣1+t1−tt​1−t1+t−t​121​∣

$${C}_{2}\rightarrow {C}_{1}+{C}_{2}$$C2​→C1​+C2​

$$=|\begin{matrix}1+t & 2 & 1 \\ 1-t & 2 & 2 \\ t & 0 & 1\end{matrix}|$$=∣1+t1−tt​220​121​∣

$$=2|\begin{matrix}1+t & 1 & 1 \\ 1-t & 1 & 2 \\ t & 0 & 1\end{matrix}|$$=2∣1+t1−tt​110​121​∣

$$=2[(1+t)-(1-t)+t]$$=2[(1+t)−(1−t)+t]

$$=2[3t]=6t$$=2[3t]=6t

Now from equation $(1)$(1) we get, $[\begin{matrix}\vec{a} & \vec{b} & \vec{c}\end{matrix}]\neq 0\Rightarrow t\neq 0$[a​b​c​]=0⇒t=0