1. Given vectors

$\vec a=\hat i+2\hat k=(1,0,2),\qquad \vec b=\hat i+\hat j+\hat k=(1,1,1),\qquad \vec c=7\hat i-3\hat j+4\hat k=(7,-3,4)$

We are given

$\vec r\times \vec b+\vec b\times \vec c=\vec 0$

and

$\vec r\cdot \vec a=0.$

We need to find $\vec r\cdot \vec c$.

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2. Simplify the cross product equation

Using anti-commutativity of cross product,

$\vec b\times \vec c = -\vec c\times \vec b.$

So,

$\vec r\times \vec b+\vec b\times \vec c=0$ $\Rightarrow \vec r\times \vec b-\vec c\times \vec b=0$ $\Rightarrow (\vec r-\vec c)\times \vec b=0.$

Hence $\vec r-\vec c$ is parallel to $\vec b$. Therefore,

$\vec r=\vec c+\lambda \vec b$

for some scalar $\lambda$.

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3. Use the dot product condition

Given

$\vec r\cdot \vec a=0$

Substitute $\vec r=\vec c+\lambda \vec b$:

$(\vec c+\lambda \vec b)\cdot \vec a=0$ $\vec c\cdot \vec a+\lambda (\vec b\cdot \vec a)=0.$

Now compute:

$\vec c\cdot \vec a=(7,-3,4)\cdot(1,0,2)=7+0+8=15$

$\vec b\cdot \vec a=(1,1,1)\cdot(1,0,2)=1+0+2=3$

So,

$15+3\lambda=0$ $\lambda=-5.$

Thus,

$\vec r=\vec c-5\vec b.$

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4. Find $\vec r\cdot \vec c$

$\vec r\cdot \vec c=(\vec c-5\vec b)\cdot \vec c$ $=\vec c\cdot \vec c-5(\vec b\cdot \vec c).$

Compute each term:

$\vec c\cdot \vec c=7^2+(-3)^2+4^2=49+9+16=74$

$\vec b\cdot \vec c=(1,1,1)\cdot(7,-3,4)=7-3+4=8$

Therefore,

$\vec r\cdot \vec c=74-5(8)=74-40=34.$

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5. Check options

The value is

$\boxed{34}$

So the correct option is C.