1. Given vectors

$\vec a=(4,-1,1), \qquad \vec b=(11,-1,1)$

We are given

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

Also,

$(2\vec a+3\vec b)\cdot \vec c=1670.$

We need to find $|\vec c|^2$.

2. Use cross product property

Recall:

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

So,

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

Hence the given equation becomes

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

Bring both to one side:

$[(\vec a+\vec b)-(2\vec a-3\vec b)]\times \vec c=\vec 0.$

That is,

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

Therefore $\vec c$ is parallel to $-\vec a+4\vec b$.

So let

$\vec c=\lambda(-\vec a+4\vec b).$

3. Compute the needed vectors

First,

$-\vec a+4\vec b=-(4,-1,1)+4(11,-1,1)=( -4,1,-1)+(44,-4,4)=(40,-3,3).$

Thus,

$\vec c=\lambda(40,-3,3).$

Now compute

$2\vec a+3\vec b=2(4,-1,1)+3(11,-1,1)=(8,-2,2)+(33,-3,3)=(41,-5,5).$

Given

$(2\vec a+3\vec b)\cdot \vec c=1670,$

so

$(41,-5,5)\cdot \lambda(40,-3,3)=1670.$

Compute dot product:

$41\cdot 40+(-5)(-3)+5\cdot 3=1640+15+15=1670.$

Hence,

$\lambda(1670)=1670 \implies \lambda=1.$

Therefore,

$\vec c=(40,-3,3).$

4. Find $|\vec c|^2$

$|\vec c|^2=40^2+(-3)^2+3^2=1600+9+9=1618.$

5. Option check

• A: $1600$ ❌
• B: $1618$ ✅
• C: $1627$ ❌
• D: $1609$ ❌

So the correct answer is:

$\boxed{1618}$