If a = i + j + k , b = j - k and c be three vectors such that a × c...

Given $\vec{a} \times \vec{c} = \vec{b}$.

Taking the dot product with $\vec{c}$ on both sides:

$(\vec{a} \times \vec{c}) \cdot \vec{c} = \vec{b} \cdot \vec{c}$

Since the scalar triple product with two identical vectors is zero, we get:

$\vec{b} \cdot \vec{c} = 0$

We need to find the value of $\vec{c} \cdot (\vec{a} - 2\vec{b})$. Expanding the dot product:

$\vec{c} \cdot (\vec{a} - 2\vec{b}) = \vec{c} \cdot \vec{a} - 2(\vec{c} \cdot \vec{b})$

It is given that $\vec{a} \cdot \vec{c} = 3$. Substituting the values:

$\vec{c} \cdot (\vec{a} - 2\vec{b}) = 3 - 2(0) = 3$

Answer: $3$