Let $$\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}$$, $$\vec{b} = \hat{i} + 2\hat{j} - 4\hat{k}$$ and $$\vec{c} = ((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$$. Then $$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$$ is equal to:

Given vectors are:

$$\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}$$

$$\vec{b} = \hat{i} + 2\hat{j} - 4\hat{k}$$

We need to find $$\vec{c} = ((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$$ and then compute $$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$$.

First, compute $$\vec{d} = \vec{a} \times \vec{b}$$ using the cross product formula:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -3 \\ 1 & 2 & -4 \\ \end{vmatrix}$$

Expanding the determinant:

$$\hat{i} \left( (1)(-4) - (-3)(2) \right) - \hat{j} \left( (-5)(-4) - (-3)(1) \right) + \hat{k} \left( (-5)(2) - (1)(1) \right)$$

Calculate each component:

• $$\hat{i}$$ component: $$(1)(-4) - (-3)(2) = -4 + 6 = 2$$
• $$\hat{j}$$ component: $$- \left[ (-5)(-4) - (-3)(1) \right] = - \left[ 20 + 3 \right] = -23$$
• $$\hat{k}$$ component: $$(-5)(2) - (1)(1) = -10 - 1 = -11$$

So, $$\vec{d} = 2\hat{i} - 23\hat{j} - 11\hat{k}$$

Next, compute $$\vec{e} = \vec{d} \times \hat{i}$$. The cross product of any vector $$\vec{u} = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$$ with $$\hat{i}$$ is given by:

$$\vec{u} \times \hat{i} = (u_z) \hat{j} + (-u_y) \hat{k}$$

Here, $$\vec{d} = 2\hat{i} - 23\hat{j} - 11\hat{k}$$, so $$u_x = 2$$, $$u_y = -23$$, $$u_z = -11$$.

Thus, $$\vec{e} = (-11) \hat{j} + (-(-23)) \hat{k} = -11\hat{j} + 23\hat{k}$$

Now, compute $$\vec{f} = \vec{e} \times \hat{i}$$ using the same formula:

$$\vec{e} = 0\hat{i} -11\hat{j} + 23\hat{k}$$, so $$u_x = 0$$, $$u_y = -11$$, $$u_z = 23$$.

Thus, $$\vec{f} = (23) \hat{j} + (-(-11)) \hat{k} = 23\hat{j} + 11\hat{k}$$

Finally, compute $$\vec{c} = \vec{f} \times \hat{i}$$:

$$\vec{f} = 0\hat{i} + 23\hat{j} + 11\hat{k}$$, so $$u_x = 0$$, $$u_y = 23$$, $$u_z = 11$$.

Thus, $$\vec{c} = (11) \hat{j} + (-23) \hat{k} = 11\hat{j} - 23\hat{k}$$

Now, compute the dot product $$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$$:

$$\vec{c} = 0\hat{i} + 11\hat{j} - 23\hat{k}$$

Dot product: $$(0)(-1) + (11)(1) + (-23)(1) = 0 + 11 - 23 = -12$$

Therefore, the value is $$-12$$, which corresponds to option A.