Let $$\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$$ be two vectors. If a vector perpendicular to both the vectors $$\vec{a} + \vec{b}$$ and $$\vec{a} - \vec{b}$$ has the magnitude 12 then one such vector is:

We have the two given vectors $$\vec a = 3\hat i + 2\hat j + 2\hat k$$ and $$\vec b = \hat i + 2\hat j - 2\hat k.$$

First we form the combinations that are mentioned in the question.

Adding, we get

$$\vec a + \vec b = (3+1)\hat i + (2+2)\hat j + (2-2)\hat k = 4\hat i + 4\hat j + 0\hat k.$$

Subtracting, we get

$$\vec a - \vec b = (3-1)\hat i + (2-2)\hat j + (2-(-2))\hat k = 2\hat i + 0\hat j + 4\hat k.$$

A vector that is perpendicular to both $$\vec a + \vec b$$ and $$\vec a - \vec b$$ must be parallel to their cross-product. We therefore recall the formula:

For any vectors $$\vec u = u_1\hat i + u_2\hat j + u_3\hat k$$ and $$\vec v = v_1\hat i + v_2\hat j + v_3\hat k,$$ their cross-product is

$$\vec u \times \vec v = \begin{vmatrix} \hat i & \hat j & \hat k\\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3 \end{vmatrix}.$$

Applying this to $$\vec u = \vec a + \vec b = 4\hat i + 4\hat j + 0\hat k$$ and $$\vec v = \vec a - \vec b = 2\hat i + 0\hat j + 4\hat k,$$ we write

$$ \vec u \times \vec v = \begin{vmatrix} \hat i & \hat j & \hat k\\ 4 & 4 & 0\\ 2 & 0 & 4 \end{vmatrix}. $$

Expanding the determinant step by step, we obtain

$$ \vec u \times \vec v = \hat i\,(4\cdot 4 - 0\cdot 0) \;-\; \hat j\,(4\cdot 4 - 0\cdot 2) \;+\; \hat k\,(4\cdot 0 - 4\cdot 2). $$

Simplifying every product inside the brackets, we have

$$ \vec u \times \vec v = \hat i\,(16 - 0) \;-\; \hat j\,(16 - 0) \;+\; \hat k\,(0 - 8). $$

Thus

$$ \vec u \times \vec v = 16\hat i - 16\hat j - 8\hat k. $$

This vector is perpendicular to both $$\vec a+\vec b$$ and $$\vec a-\vec b.$

Now we calculate its magnitude, using the formula $$|$$\vec$$ r| = $$\sqrt{r_x^2 + r_y^2 + r_z^2}$$.$$ Substituting $$r_x = 16,\; r_y = -16,\; r_z = -8,$$ we get

$$ |$$\vec$$ u $$\times$$ $$\vec$$ v| = $$\sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576}$$ = 24. $$

The question demands a perpendicular vector of magnitude $$12,$$ which is exactly one-half of $$24.$$ Therefore we divide each component of $$16$$\hat$$ i - 16$$\hat$$ j - 8$$\hat$$ k$$ by $$2.$$ We get

$$ $$\frac{1}{2}$$\bigl(16$$\hat$$ i - 16$$\hat$$ j - 8$$\hat$$ k\bigr) = 8$$\hat$$ i - 8$$\hat$$ j - 4$$\hat$$ k. $$

This new vector has magnitude

$$ $$\sqrt{8^2 + (-8)^2 + (-4)^2} = \sqrt{64 + 64 + 16} = \sqrt{144}$$ = 12, $$

exactly as required.

Factoring out a common multiple of $$4,$$ we write

$$ 8$$\hat$$ i - 8$$\hat$$ j - 4$$\hat$$ k = 4\,(2$$\hat$$ i - 2$$\hat$$ j - $$\hat$$ k). $$

Looking at the options, this matches Option B, namely $$4(2$$\hat$$ i - 2$$\hat$$ j - $$\hat$$ k).$$

Hence, the correct answer is Option B.