Given vectors $\vec a = 2\hat i+5\hat j-\hat k, \qquad \vec b = 2\hat i-2\hat j+2\hat k$ Let $\vec c = x\hat i+y\hat j+z\hat k.$

Use the vector equation $(\vec c+\hat i)\times(\vec a+\vec b+\hat i)=\vec a\times(\vec c+\hat i).$

First compute $\vec a+\vec b+\hat i=(2+2+1)\hat i+(5-2)\hat j+(-1+2)\hat k=5\hat i+3\hat j+\hat k.$

So the equation becomes $(\vec c+\hat i)\times(5\hat i+3\hat j+\hat k)=\vec a\times(\vec c+\hat i).$

Bring all terms to one side Using anti-commutativity of cross product, $\vec a\times(\vec c+\hat i)=-(\vec c+\hat i)\times \vec a.$ Hence $(\vec c+\hat i)\times(5\hat i+3\hat j+\hat k)+(\vec c+\hat i)\times\vec a=\vec 0.$

Factor out $(\vec c+\hat i)\times$: $(\vec c+\hat i)\times\big((5\hat i+3\hat j+\hat k)+\vec a\big)=\vec 0.$

Now $(5\hat i+3\hat j+\hat k)+\vec a=(5,3,1)+(2,5,-1)=(7,8,0).$

Therefore, $(\vec c+\hat i)\times(7\hat i+8\hat j)=\vec 0.$

Interpret the cross product zero condition If cross product is zero, the vectors are parallel: $\vec c+\hat i=\lambda(7\hat i+8\hat j).$

So $\vec c=(7\lambda-1)\hat i+8\lambda\hat j+0\hat k.$

Hence, $x=7\lambda-1,\quad y=8\lambda,\quad z=0.$

Use the dot product condition Given $\vec a\cdot \vec c=-29.$ Since $\vec a=(2,5,-1), \quad \vec c=(7\lambda-1,8\lambda,0),$ we get $2(7\lambda-1)+5(8\lambda)+(-1)(0)=-29.$ $14\lambda-2+40\lambda=-29$ $54\lambda-2=-29$ $54\lambda=-27$ $\lambda=-\frac12.$

Therefore, $\vec c=\left(7\left(-\frac12\right)-1,\,8\left(-\frac12\right),\,0\right)=\left(-\frac72-1,-4,0\right)=\left(-\frac92,-4,0\right).$

Compute the required dot product We need $\vec c\cdot(-2\hat i+\hat j+\hat k).$

So $\left(-\frac92,-4,0\right)\cdot(-2,1,1)$ $=\left(-\frac92\right)(-2)+(-4)(1)+0(1)$ $=9-4=5.$

Compare with options $5$ corresponds to Option C.