If $$\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}, \vec{b} = 3\hat{i} + 3\hat{j} + \hat{k}$$ and $$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$$ are coplanar vectors and $$\vec{a} \cdot \vec{c} = 5, \vec{b} \perp \vec{c}$$, then $$122(c_1 + c_2 + c_3)$$ is equal to ______

Given $$\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$$, $$\vec{b} = 3\hat{i} + 3\hat{j} + \hat{k}$$, and $$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$$.

Conditions: $$\vec{a}, \vec{b}, \vec{c}$$ are coplanar, $$\vec{a} \cdot \vec{c} = 5$$, and $$\vec{b} \perp \vec{c}$$.

Since the vectors are coplanar: $$\vec{c} = \lambda \vec{a} + \mu \vec{b}$$ for some scalars $$\lambda, \mu$$.

$$\vec{c} = (2\lambda + 3\mu)\hat{i} + (\lambda + 3\mu)\hat{j} + (3\lambda + \mu)\hat{k}$$

From $$\vec{b} \perp \vec{c}$$: $$\vec{b} \cdot \vec{c} = 0$$

$$3(2\lambda + 3\mu) + 3(\lambda + 3\mu) + 1(3\lambda + \mu) = 0$$

$$6\lambda + 9\mu + 3\lambda + 9\mu + 3\lambda + \mu = 0$$

$$12\lambda + 19\mu = 0 \implies \lambda = -\frac{19\mu}{12}$$

From $$\vec{a} \cdot \vec{c} = 5$$:

$$2(2\lambda + 3\mu) + 1(\lambda + 3\mu) + 3(3\lambda + \mu) = 5$$

$$4\lambda + 6\mu + \lambda + 3\mu + 9\lambda + 3\mu = 5$$

$$14\lambda + 12\mu = 5$$

Substituting $$\lambda = -\frac{19\mu}{12}$$:

$$14 \cdot \left(-\frac{19\mu}{12}\right) + 12\mu = 5$$

$$-\frac{266\mu}{12} + 12\mu = 5$$

$$-\frac{133\mu}{6} + 12\mu = 5$$

$$\frac{-133\mu + 72\mu}{6} = 5$$

$$\frac{-61\mu}{6} = 5$$

$$\mu = -\frac{30}{61}$$

$$\lambda = -\frac{19}{12} \cdot \left(-\frac{30}{61}\right) = \frac{19 \times 30}{12 \times 61} = \frac{570}{732} = \frac{95}{122}$$

Now computing $$c_1 + c_2 + c_3$$:

$$c_1 + c_2 + c_3 = (2\lambda + 3\mu) + (\lambda + 3\mu) + (3\lambda + \mu) = 6\lambda + 7\mu$$

$$= 6 \cdot \frac{95}{122} + 7 \cdot \left(-\frac{30}{61}\right) = \frac{570}{122} - \frac{210}{61} = \frac{570}{122} - \frac{420}{122} = \frac{150}{122} = \frac{75}{61}$$

$$122(c_1 + c_2 + c_3) = 122 \times \frac{75}{61} = 2 \times 75 = 150$$

Hence the answer is $$\boxed{150}$$.