If $\vec{a}= 2\hat{i} -\hat{j} + \hat{k}, \vec{b} =\hat{i} + \hat{j} - 2\hat{k}, \vec{c} =\hat{i} + 3\hat{j} - \hat{k}, $ such that $\vec{a} $ is perpendicular to $\left(\lambda \vec{b} + \vec{c} \right), $ then the value of $\lambda $ is :

-2

The correct answer is Option (4) → -2

$\vec a⊥(λ\vec b+\vec c)$

So $\vec a.(λ\vec b+\vec c)=0$

So $(2\hat i-\hat j+\hat k)\left((λ+1)\hat i+(λ+3)\hat j+(-2λ-1)\hat k\right)$

$⇒2λ+2-λ-3λ-2λ-1=0$

$⇒-2=λ$