If $$|\vec{c}|^2 = 60$$ and $$\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$$, then a value of $$\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$$ is:

We are given two conditions: $$|\vec{c}|^2 = 60$$ and $$\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$$. We need to find $$\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$$.

First, since the cross product $$\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}$$, the vectors $$\vec{c}$$ and $$\hat{i} + 2\hat{j} + 5\hat{k}$$ are parallel. Therefore, $$\vec{c}$$ is a scalar multiple of $$\hat{i} + 2\hat{j} + 5\hat{k}$$. Let $$\vec{c} = k (\hat{i} + 2\hat{j} + 5\hat{k})$$, where $$k$$ is a scalar. So, $$\vec{c} = k\hat{i} + 2k\hat{j} + 5k\hat{k}$$.

Now, we know $$|\vec{c}|^2 = 60$$. The magnitude squared is the dot product of $$\vec{c}$$ with itself:

$$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (k\hat{i} + 2k\hat{j} + 5k\hat{k}) \cdot (k\hat{i} + 2k\hat{j} + 5k\hat{k}) = k \cdot k + (2k) \cdot (2k) + (5k) \cdot (5k) = k^2 + 4k^2 + 25k^2 = 30k^2.$$

Given $$|\vec{c}|^2 = 60$$, we have:

$$30k^2 = 60.$$

Dividing both sides by 30:

$$k^2 = 2.$$

So, $$k = \sqrt{2}$$ or $$k = -\sqrt{2}$$.

Next, we compute $$\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$$. Let $$\vec{d} = -7\hat{i} + 2\hat{j} + 3\hat{k}$$. Then:

$$\vec{c} \cdot \vec{d} = (k\hat{i} + 2k\hat{j} + 5k\hat{k}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k}).$$

The dot product is:

$$(k) \cdot (-7) + (2k) \cdot (2) + (5k) \cdot (3) = -7k + 4k + 15k.$$

Combining like terms:

$$-7k + 4k = -3k,$$

$$-3k + 15k = 12k.$$

So, $$\vec{c} \cdot \vec{d} = 12k$$.

Substituting the values of $$k$$:

If $$k = \sqrt{2}$$, then $$\vec{c} \cdot \vec{d} = 12 \cdot \sqrt{2} = 12\sqrt{2}$$.

If $$k = -\sqrt{2}$$, then $$\vec{c} \cdot \vec{d} = 12 \cdot (-\sqrt{2}) = -12\sqrt{2}$$.

The problem asks for "a value" of the dot product. The options provided are:

A. $$4\sqrt{2}$$

B. 12

C. 24

D. $$12\sqrt{2}$$

Among the possible values, $$12\sqrt{2}$$ is present in the options as choice D. The value $$-12\sqrt{2}$$ is not listed. Therefore, we select $$12\sqrt{2}$$.

Hence, the correct answer is Option D.