Let $$\vec{a} = 4\hat{i} - \hat{j} + \hat{k}$$, $$\vec{b} = 11\hat{i} - \hat{j} + \hat{k}$$ and $$\vec{c}$$ be a vector such that $$(\vec{a} + \vec{b}) \times \vec{c} = \vec{c} \times (-2\vec{a} + 3\vec{b})$$. If $$(2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670$$, then $$|\vec{c}|^2$$ is equal to :

Given $$\vec{a} = 4\hat{i} - \hat{j} + \hat{k}$$, $$\vec{b} = 11\hat{i} - \hat{j} + \hat{k}$$, and $$(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times(-2\vec{a}+3\vec{b})$$.

We use the fact that

$$\vec{c} \times(-2\vec{a}+3\vec{b}) = -(-2\vec{a}+3\vec{b}) \times \vec{c}$$.

Hence

$$ (\vec{a}+\vec{b}) \times \vec{c} + (-2\vec{a}+3\vec{b}) \times \vec{c} = \vec{0} $$

Applying the distributive property of the cross product leads to

$$ ((\vec{a}+\vec{b}) + (-2\vec{a}+3\vec{b})) \times \vec{c} = \vec{0} $$

which simplifies to

$$ (-\vec{a}+4\vec{b}) \times \vec{c} = \vec{0}. $$

Therefore, $$\vec{c}$$ must be parallel to $$(-\vec{a}+4\vec{b})$$.

Substituting the component forms gives

$$ -\vec{a} + 4\vec{b} = -(4\hat{i}-\hat{j}+\hat{k}) + 4(11\hat{i}-\hat{j}+\hat{k}) = -4\hat{i}+\hat{j}-\hat{k}+44\hat{i}-4\hat{j}+4\hat{k} = 40\hat{i}-3\hat{j}+3\hat{k}. $$

Thus we can write $$\vec{c} = t(40\hat{i}-3\hat{j}+3\hat{k})$$ for some scalar $$t$$.

Using the condition $$(2\vec{a}+3\vec{b}) \cdot \vec{c} = 1670$$ we compute

$$ 2\vec{a} + 3\vec{b} = 2(4\hat{i}-\hat{j}+\hat{k}) + 3(11\hat{i}-\hat{j}+\hat{k}) = 8\hat{i}-2\hat{j}+2\hat{k}+33\hat{i}-3\hat{j}+3\hat{k} = 41\hat{i}-5\hat{j}+5\hat{k}. $$

Substitution into the dot product yields

$$ (41\hat{i}-5\hat{j}+5\hat{k}) \cdot t(40\hat{i}-3\hat{j}+3\hat{k}) = 1670. $$

This simplifies to

$$ t(41 \times 40 + (-5)(-3) + 5 \times 3) = 1670, $$

so

$$ t(1640 + 15 + 15) = 1670, $$

giving

$$ t \times 1670 = 1670 \implies t = 1. $$

Therefore $$\vec{c} = 40\hat{i}-3\hat{j}+3\hat{k}$$ and its squared magnitude is

$$ |\vec{c}|^2 = 40^2 + 3^2 + 3^2 = 1600 + 9 + 9 = 1618. $$

The correct answer is Option 2: 1618.