1. Let $\vec a=(2,1,3),\quad \vec b=(3,3,1),\quad \vec c=(c_1,c_2,c_3).$

Given:

• $\vec a,\vec b,\vec c$ are coplanar
• $\vec a\cdot \vec c=5$
• $\vec b\perp \vec c \Rightarrow \vec b\cdot \vec c=0$

2. Since $\vec a,\vec b,\vec c$ are coplanar and $\vec a,\vec b$ are not parallel, we can write $\vec c = x\vec a + y\vec b.$

So, $\vec c = x(2,1,3)+y(3,3,1).$

3. Use the dot product conditions.

First compute needed dot products: $\vec a\cdot \vec a = 2^2+1^2+3^2=14,$ $\vec a\cdot \vec b = 2\cdot 3+1\cdot 3+3\cdot 1=12,$ $\vec b\cdot \vec b = 3^2+3^2+1^2=19.$

Now, $\vec a\cdot \vec c = \vec a\cdot (x\vec a+y\vec b)=14x+12y=5 \quad ...(1)$

and $\vec b\cdot \vec c = \vec b\cdot (x\vec a+y\vec b)=12x+19y=0 \quad ...(2)$

4. Solve equations (1) and (2):

From (2), $12x=-19y \Rightarrow x=-\frac{19}{12}y.$

Substitute into (1): $14\left(-\frac{19}{12}y\right)+12y=5.$

$-\frac{266}{12}y+12y=5$ $-\frac{133}{6}y+\frac{72}{6}y=5$ $-\frac{61}{6}y=5$ $y=-\frac{30}{61}.$

Then $x=-\frac{19}{12}\left(-\frac{30}{61}\right)=\frac{95}{122}.$

5. Now find $\vec c$: $\vec c=x\vec a+y\vec b.$

So, $c_1+c_2+c_3 = x(2+1+3)+y(3+3+1).$

That is, $c_1+c_2+c_3 = 6x+7y.$

Substitute $x=\frac{95}{122}$ and $y=-\frac{30}{61}=-\frac{60}{122}$: $c_1+c_2+c_3 = 6\cdot \frac{95}{122}+7\cdot \left(-\frac{60}{122}\right).$

$=\frac{570}{122}-\frac{420}{122}=\frac{150}{122}.$

Therefore, $122(c_1+c_2+c_3)=150.$

6. Final answer: $\boxed{150}$