Let $$\vec{a} = 3\hat{i} + 2\hat{j} + \hat{k}$$, $$\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$$ and $$\vec{c}$$ be a vector such that $$(\vec{a} + \vec{b}) \times \vec{c} = 2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k}$$ and $$(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = -3$$. Then $$|\vec{c}|^2$$ is equal to

We start by noting $$\vec{a} = 3\hat{i} + 2\hat{j} + \hat{k}$$ and $$\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$$ and the requirement to find $$\vec{c}$$ such that:

$$ (\vec{a}+\vec{b}) \times \vec{c} = 2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k} \quad \cdots(1) $$

$$ (\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c} = -3 \quad \cdots(2) $$

Next, we compute $$\vec{a}+\vec{b} = 5\hat{i} + \hat{j} + 4\hat{k}$$ and $$\vec{a}-\vec{b}+\hat{i} = (3-2+1)\hat{i} + (2+1)\hat{j} + (1-3)\hat{k} = 2\hat{i} + 3\hat{j} - 2\hat{k}$$.

Substituting into the cross product formula gives $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & -1 & 3 \end{vmatrix} = \hat{i}(6+1) - \hat{j}(9-2) + \hat{k}(-3-4) = 7\hat{i} - 7\hat{j} - 7\hat{k}$$.

This gives us $$2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k} = 14\hat{i} - 14\hat{j} - 14\hat{k} + 24\hat{j} - 6\hat{k} = 14\hat{i} + 10\hat{j} - 20\hat{k}$$.

We let $$\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$$ and compute

$$ (\vec{a}+\vec{b}) \times \vec{c} = (5\hat{i}+\hat{j}+4\hat{k}) \times (x\hat{i}+y\hat{j}+z\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix} = \hat{i}(z-4y) - \hat{j}(5z-4x) + \hat{k}(5y-x)$$.

Equating this to $$14\hat{i} + 10\hat{j} - 20\hat{k}$$ yields the system

$$ z - 4y = 14 \quad \cdots(3) $$

$$ 4x - 5z = 10 \quad \cdots(4) $$

$$ 5y - x = -20 \quad \cdots(5) $$

From equation (2) we also have $$2x + 3y - 2z = -3 \quad \cdots(6)$$.

Solving (3) for z gives $$z = 14 + 4y$$ and solving (5) for x gives $$x = 5y + 20$$.

Substituting into (4) gives $$4(5y+20) - 5(14+4y) = 10$$, which simplifies to $$20y + 80 - 70 - 20y = 10$$ and holds identically.

Using equation (6) then yields $$2(5y+20) + 3y - 2(14+4y) = -3$$ or $$10y + 40 + 3y - 28 - 8y = -3$$ leading to $$5y + 12 = -3$$ and hence $$y = -3$$.

Therefore, $$x = 5(-3)+20 = 5$$ and $$z = 14+4(-3) = 2$$, so $$\vec{c} = 5\hat{i} - 3\hat{j} + 2\hat{k}$$.

Finally, $$|\vec{c}|^2 = 25 + 9 + 4 = 38$$.

The answer is 38.