Let $$\vec{a} = \hat{i} + \hat{j} + \hat{k}$$ and $$\vec{b} = \hat{j} - \hat{k}$$. If $$\vec{c}$$ is a vector such that $$\vec{a} \times \vec{c} = \vec{b}$$ and $$\vec{a} \cdot \vec{c} = 3$$, then $$\vec{a} \cdot (\vec{b} \times \vec{c})$$ is equal to:

We have the vectors

$$\vec a = \hat i + \hat j + \hat k, \qquad \vec b = \hat j - \hat k.$$

Let us assume

$$\vec c = x\,\hat i + y\,\hat j + z\,\hat k.$$

The first condition is the cross-product relation

$$\vec a \times \vec c = \vec b.$$
For two vectors $$\vec p = (p_x,\,p_y,\,p_z),\qquad \vec q = (q_x,\,q_y,\,q_z),$$ the formula for the cross product is $$\vec p \times \vec q = \bigl(p_yq_z - p_zq_y\bigr)\hat i - \bigl(p_xq_z - p_zq_x\bigr)\hat j + \bigl(p_xq_y - p_yq_x\bigr)\hat k.$$
Applying this to $$\vec p = \vec a =(1,1,1),\quad \vec q = \vec c =(x,y,z),$$ we obtain

$$ \vec a \times \vec c = \bigl(1\cdot z - 1\cdot y\bigr)\hat i - \bigl(1\cdot z - 1\cdot x\bigr)\hat j + \bigl(1\cdot y - 1\cdot x\bigr)\hat k = (z-y)\,\hat i -(z-x)\,\hat j +(y-x)\,\hat k. $$

Since this must equal $$\vec b = 0\,\hat i + 1\,\hat j -1\,\hat k,$$ we equate the components:

$$ \begin{aligned} z-y &= 0, \\ -(z-x) &= 1, \\ y-x &= -1. \end{aligned} $$

From the first equation $$z=y.$$ Substituting $$z=y$$ in the second gives $$-(y-x)=1 \; \Longrightarrow \; y-x=-1.$$ The third equation is exactly the same, so the system is self-consistent.

Thus we can write

$$ y = x-1, \qquad z = x-1. $$

Now use the dot-product condition

$$\vec a \cdot \vec c = 3.$$
For vectors $$\vec p=(p_x,p_y,p_z)$$ and $$\vec q=(q_x,q_y,q_z)$$ the dot product is $$\vec p \cdot \vec q = p_xq_x + p_yq_y + p_zq_z.$$ Therefore

$$ \vec a \cdot \vec c = 1\cdot x + 1\cdot y + 1\cdot z = x + y + z = x + (x-1) + (x-1) = 3x - 2. $$

Setting this equal to 3, we get

$$3x - 2 = 3 \;\Longrightarrow\; 3x = 5 \;\Longrightarrow\; x = \frac53.$$

Hence

$$ y = x-1 = \frac53 - 1 = \frac23, \qquad z = \frac23. $$

So

$$\vec c = \frac53\,\hat i + \frac23\,\hat j + \frac23\,\hat k.$$

We now find the required scalar triple product $$\vec a \cdot (\vec b \times \vec c).$$ Again using the cross-product formula, first compute $$\vec b \times \vec c.$$

Write the components explicitly: $$\vec b = (0,1,-1), \qquad \vec c = \Bigl(\frac53,\frac23,\frac23\Bigr).$$

Then

$$ \vec b \times \vec c = \Bigl(1\cdot\frac23 - (-1)\cdot\frac23\Bigr)\hat i - \Bigl(0\cdot\frac23 - (-1)\cdot\frac53\Bigr)\hat j + \Bigl(0\cdot\frac23 - 1\cdot\frac53\Bigr)\hat k = \frac43\,\hat i -\frac53\,\hat j -\frac53\,\hat k. $$

Finally take the dot product with $$\vec a = (1,1,1):$$

$$ \vec a \cdot (\vec b \times \vec c) = 1\left(\frac43\right) + 1\left(-\frac53\right) + 1\left(-\frac53\right) = \frac43 - \frac53 - \frac53 = \frac{4 - 5 - 5}{3} = \frac{-6}{3} = -2. $$

So, the value of $$\vec a \cdot (\vec b \times \vec c)$$ is $$-2.$$

Hence, the correct answer is Option B.