1. Given vectors

$\vec a=2\hat i+\lambda_1\hat j+3\hat k$ $\vec b=4\hat i+(3-\lambda_2)\hat j+6\hat k$ $\vec c=3\hat i+6\hat j+(\lambda_3-1)\hat k$

We are given:

• $\vec b=2\vec a$
• $\vec a \perp \vec c$

2. Use the condition $\vec b=2\vec a$

First compute:

$2\vec a=2(2\hat i+\lambda_1\hat j+3\hat k)=4\hat i+2\lambda_1\hat j+6\hat k$

Since

$\vec b=4\hat i+(3-\lambda_2)\hat j+6\hat k,$

comparing the $\hat j$ components,

$3-\lambda_2=2\lambda_1$

So,

$\lambda_2=3-2\lambda_1$

3. Use the perpendicular condition $\vec a\cdot \vec c=0$

$\vec a\cdot \vec c=(2)(3)+(\lambda_1)(6)+(3)(\lambda_3-1)=0$

$6+6\lambda_1+3\lambda_3-3=0$

$3+6\lambda_1+3\lambda_3=0$

Divide by $3$:

$1+2\lambda_1+\lambda_3=0$

So,

$\lambda_3=-1-2\lambda_1$

4. Now check the options

We need values satisfying both:

$\lambda_2=3-2\lambda_1, \qquad \lambda_3=-1-2\lambda_1$

Option A: $(1,5,1)$

If $\lambda_1=1$,

$\lambda_2=3-2(1)=1 \neq 5$

So A is incorrect.

Option B: $(1,3,1)$

If $\lambda_1=1$,

$\lambda_2=1 \neq 3$

So B is incorrect.

Option C: $\left(-\frac12,4,0\right)$

If $\lambda_1=-\frac12$,

$\lambda_2=3-2\left(-\frac12\right)=3+1=4$

and

$\lambda_3=-1-2\left(-\frac12\right)=-1+1=0$

Both match. So C is correct.

Option D: $\left(\frac12,4,-2\right)$

If $\lambda_1=\frac12$,

$\lambda_2=3-2\left(\frac12\right)=2 \neq 4$

So D is incorrect.

5. Final answer

The possible value is

$\boxed{\left(-\frac12,4,0\right)}$

So the correct option is C.