The plane, passing through the points $(0, –1, 2)$ and $(–1, 2, 1)$ and parallel to the line passing through $(5, 1, –7)$ and $(1, –1, –1),$also passes through the point

Let $A\left(0,-1,2\right) \& B\left(-1,2,1\right)$ and given plane is parallel to line passing through $\left(5,1,-7\right) \& \left(1,-1,-1\right)$

So, normal vector to the plane is given by,

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1-0& 2+1& 1-2\\ 1-5& -1-1& -1+7\end{array}\right|$

$\Rightarrow \overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 3& -1\\ -4& -2& 6\end{array}\right|=\hat{i}\left(16\right)-\hat{j}\left(-10\right)+\hat{k}\left(14\right)$

$\Rightarrow \overrightarrow{n}=16\hat{i}+10\hat{j}+14\hat{k}$

$\therefore$Equation of plane

$16(x – 0) + 10(y + 1) + 14 (z – 2) = 0$

$\Rightarrow 8x+5y+7z=9$

Now satisfying the points from option in the plane we get, 

$\left(-2,5,0\right)$ only point which satisfy the plane.