1. Given condition on cross products

We have

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

Let

$$\vec a=(2,-1,3),\quad \vec b=(3,-5,1),\quad \vec c=(x,y,z).$$

From

$$\vec a\times \vec c=\vec a\times \vec b$$

we get

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

Hence $\vec c-\vec b$ is parallel to $\vec a$, so

$$\vec c=\vec b+\lambda \vec a.$$

Also, from

$$\vec a\times \vec b=\vec c\times \vec b$$

we get

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

Hence $\vec a-\vec c$ is parallel to $\vec b$, so

$$\vec c=\vec a-\mu \vec b.$$

Therefore $\vec c$ lies on both lines:

$$\vec c=\vec b+\lambda \vec a=\vec a-\mu \vec b.$$

A better way is to use the first form

$$\vec c=\vec b+\lambda \vec a.$$

and impose the second condition.

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2. Use the equality $\vec c\times \vec b=\vec a\times \vec b$

If

$$\vec c=\vec b+\lambda \vec a,$$

then

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

Given that

$$\vec c\times \vec b=\vec a\times \vec b,$$

we get

$$\lambda(\vec a\times \vec b)=\vec a\times \vec b.$$

Since $\vec a$ and $\vec b$ are not parallel, $\vec a\times \vec b\neq \vec 0$, so

$$\lambda=1.$$

Thus

$$\vec c=\vec a+\vec b.$$

But this seems to make $\vec c$ unique, whereas the question asks for a maximum value. So let us re-check carefully.

Actually, from

$$\vec a\times \vec c=\vec c\times \vec b$$

we have

$$\vec a\times \vec c-\vec c\times \vec b=0.$$

Using $\vec c\times \vec b=-\vec b\times \vec c$,

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

Hence

$$\vec c \parallel (\vec a+\vec b).$$

So let

$$\vec c=t(\vec a+\vec b).$$

This is the correct interpretation.

Now use

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

Then

$$\vec a\times \bigl(t(\vec a+\vec b)\bigr)=t(\vec a\times \vec a+\vec a\times \vec b)=t(\vec a\times \vec b).$$

So

$$t(\vec a\times \vec b)=\vec a\times \vec b.$$

Again, since $\vec a\times \vec b\neq 0$, we get

$$t=1.$$

Hence

$$\vec c=\vec a+\vec b.$$

Still unique. Therefore the intended condition must allow two possibilities through the dot-product equation; let us compute directly from the vector equation in a more general way.

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3. Solve the cross-product equalities systematically

Given

$$\vec a\times \vec c=\vec a\times \vec b$$

so

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

Also,

$$\vec a\times \vec b=\vec c\times \vec b$$

so

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

Equating,

$$\vec b+\lambda \vec a=\vec a+\mu \vec b$$ $$(\lambda-1)\vec a+(1-\mu)\vec b=0.$$

Since $\vec a,\vec b$ are linearly independent, their coefficients must vanish:

$$\lambda=1,\quad \mu=1.$$

Thus

$$\boxed{\vec c=\vec a+\vec b}.$$

So

$$\vec c=(2,-1,3)+(3,-5,1)=(5,-6,4).$$

Then

$$|\vec c|^2=5^2+(-6)^2+4^2=25+36+16=77.$$

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4. Check the dot-product condition

Now verify whether

$$(\vec a+\vec c)\cdot(\vec b+\vec c)=168.$$

Since $\vec c=\vec a+\vec b$,

$$\vec a+\vec c=2\vec a+\vec b, \qquad \vec b+\vec c=\vec a+2\vec b.$$

So

$$(2\vec a+\vec b)\cdot(\vec a+2\vec b) =2|\vec a|^2+5\vec a\cdot\vec b+2|\vec b|^2.$$

Compute:

$$|\vec a|^2=2^2+(-1)^2+3^2=14,$$ $$|\vec b|^2=3^2+(-5)^2+1^2=35,$$ $$\vec a\cdot\vec b=2\cdot3+(-1)(-5)+3\cdot1=6+5+3=14.$$

Therefore,

$$2(14)+5(14)+2(35)=28+70+70=168.$$

The condition is satisfied.

Thus the only possible value is

$$\boxed{77}.$$

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5. Option check

• A: $77$ ✅
• B: $154$ ❌
• C: $308$ ❌
• D: $462$ ❌

So the correct option is A.