Question

Let $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\overrightarrow{ a }+(\overrightarrow{ b } \times \overrightarrow{ c })=\overrightarrow{0}$ and $\overrightarrow{ b } \cdot \overrightarrow{ c }=5$. Then, the value of $3(\vec{c} \cdot \vec{a})$ is equal to ______

Answer 1

$ \vec{a} + \vec{b} \times \vec{c} =0 $
$ \vec{a} \times \vec{b} +| \vec{b} |^{2} \vec{c} -5 \vec{b} =0$
It gives $\vec{c}=\frac{1}{3}(10 \hat{i}+3 \hat{ j }+2 \hat{ k })$
so $3 \vec{a}. \vec{c} =10$
But it does not satisfy $\vec{a} + \vec{b} \times \vec{c} =0$.
This question has data error.
Alternate (Explanation) :
According to given $a$ & $b$
$a \cdot b =1-2+3=2 \ldots$ (i)
but given equation
$\vec{a} =-( \vec{b} \times \vec{c} )$
$\Rightarrow \vec{a} \perp \vec{b} \Rightarrow \vec{a} \cdot \vec{b} =0$
which contradicts.