1. Given vectors

$$\vec a = -5\hat i + \hat j - 3\hat k, \qquad \vec b = \hat i + 2\hat j - 4\hat k$$

We need

$$\vec c = ((((\vec a \times \vec b) \times \hat i) \times \hat i) \times \hat i)$$

and then compute

$$\vec c \cdot (-\hat i + \hat j + \hat k).$$

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2. Find $\vec a \times \vec b$

Using the determinant:

$$\vec a \times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k \\ -5 & 1 & -3 \\ 1 & 2 & -4 \end{vmatrix}$$ $$= \hat i\,(1\cdot(-4)-(-3)\cdot2) - \hat j\,((-5)\cdot(-4)-(-3)\cdot1) + \hat k\,((-5)\cdot2-1\cdot1)$$ $$= \hat i(-4+6) - \hat j(20+3) + \hat k(-10-1)$$ $$= 2\hat i - 23\hat j - 11\hat k$$

So,

$$\vec p = \vec a \times \vec b = 2\hat i - 23\hat j - 11\hat k.$$

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3. Compute successive cross products with $\hat i$

Let

$$\vec p = (2,-23,-11).$$

For any vector $(x,y,z)$,

$$(x,y,z) \times \hat i = (0,z,-y).$$

First cross product

$$\vec p \times \hat i = (0,-11,23)$$

So,

$$\vec p_1 = -11\hat j + 23\hat k.$$

Second cross product

$$\vec p_1 \times \hat i = (0,23,11)$$

So,

$$\vec p_2 = 23\hat j + 11\hat k.$$

Third cross product

$$\vec p_2 \times \hat i = (0,11,-23)$$

Hence,

$$\vec c = 11\hat j - 23\hat k.$$

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4. Now compute the dot product

$$\vec c \cdot (-\hat i + \hat j + \hat k) = (0,11,-23)\cdot(-1,1,1)$$ $$= 0\cdot(-1) + 11\cdot1 + (-23)\cdot1$$ $$= 11 - 23 = -12$$

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5. Compare with options

The value is

$$\boxed{-12}$$

So the correct option is:

A.