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Point $A(1,-1,0)$ lies on the plane $2 x+y+6 z=1$ thus $A\equiv A'$
From $B(-1,0,1)$ we find the perpendicular line parametric equations $$(x=-1+2t,y=t,z=1+6t)$$ $$2 (-1+2t)+t+6(1+6t)=1\to t=-\frac{3}{41}\to B'=\left(-\frac{47}{41},-\frac{3}{41},\frac{23}{41}\right)$$ and finally $$A'B'=\sqrt{\frac{237}{41}}\approx 2.4$$
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If we name the given points as $A(1, -1, 0)$ and $B(-1,0,1)$, then,
$ \vec {AB} = (-2, 1, 1), |\vec {AB}| = \sqrt6$
Equation of plane is $2x+y+6z = 1$. So normal vector to the plane, $\vec{n} = (2,1,6)$ and $|\vec n| = \sqrt{41}$. So unit normal vector, $\hat{n} = \frac{1}{\sqrt {41}}(2,1,6)$.
Now projection of $\vec{AB}$ in the direction of normal vector to the plane is $|AB| \cos\theta$ and onto the plane is $|AB| \sin\theta$ where $\theta$ is the angle between the normal vector and $\vec{AB}$.
Projection $P_{n}$ of $\vec {AB}$ in the direction of normal vector can be simply found using dot product. $P_{n} = \vec {AB} \cdot \hat{n} = \frac{3}{\sqrt {41}}$.
Then the projection onto the plane is $P = \sqrt{|AB|^2 - P_n^2} = \sqrt{\frac{237}{41}}$
Hint: For answering the problem, find the direction vector of the line joining the points, then find the angle between the line and the normal to the plane and if that angle is $\theta$, what would be the angle between the line and the plane?
Hence what would be the projection length?