1. Given vectors

$\vec a = 6\hat i+9\hat j+12\hat k, \qquad \vec b = \alpha \hat i+11\hat j-2\hat k$

Let

$\vec c = x\hat i+y\hat j+z\hat k.$

We are given:

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

2. Use the cross product condition

From

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

we get

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

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

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

for some scalar $\lambda$.

Therefore,

$\vec c=(\alpha+6\lambda)\hat i+(11+9\lambda)\hat j+(-2+12\lambda)\hat k.$

So,

$x=\alpha+6\lambda,\quad y=11+9\lambda,\quad z=-2+12\lambda.$

3. Apply the dot product conditions

We know:

$\vec a\cdot \vec c=-12.$

Now,

$\vec a\cdot \vec c=6x+9y+12z.$

Substitute $x,y,z$:

$6(\alpha+6\lambda)+9(11+9\lambda)+12(-2+12\lambda)=-12.$

$6\alpha+36\lambda+99+81\lambda-24+144\lambda=-12$

$6\alpha+261\lambda+75=-12$

$6\alpha+261\lambda=-87 \qquad (1)$

Next,

$\vec c\cdot(\hat i-2\hat j+\hat k)=5.$

So,

$x-2y+z=5.$

Substitute:

$(\alpha+6\lambda)-2(11+9\lambda)+(-2+12\lambda)=5$

$\alpha+6\lambda-22-18\lambda-2+12\lambda=5$

$\alpha-24=5$

$\alpha=29.$

4. Find $\lambda$

Put $\alpha=29$ in equation (1):

$6(29)+261\lambda=-87$

$174+261\lambda=-87$

$261\lambda=-261$

$\lambda=-1.$

5. Find vector $\vec c$

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

Since $\alpha=29$,

$\vec b=29\hat i+11\hat j-2\hat k.$

Thus,

$\vec c=(29-6)\hat i+(11-9)\hat j+(-2-12)\hat k$

$\vec c=23\hat i+2\hat j-14\hat k.$

6. Required value

We need:

$\vec c\cdot(\hat i+\hat j+\hat k)=23+2-14=11.$

Final Answer

$\boxed{11}$