A related problem. If you know the normal vector $n=(n_1,n_2,n_3)$ to a plane and a point $p=(x_0,y_0,z_0)$ lies in the plane, then we can find the equation of the plane as
$$ n.(X-p)=0 \,,$$
where $X=(x,y,z)$ an arbitrary point lies in the plane. The point is not a problem, since you have three of them $p_1=(0,0,0)\,,p_2=(1,2,-1)\,, p_3=(0,1,1)$. The task is how to find the normal vector to the plane. I believe, you have studied the cross product of two vectors and you know the fact that the cross product of two vectors is a vector perpendicular to the plane that contains these two vectors.
Now, since you have three points, you can form two vectors
$$ v_1=p_2-p_1 \,, \quad v_2 = p_3-p_1 \,.$$
Once you form $v_1$ and $v_2$ you can find the normal to the plane as
$$ n = v_1 \times v_2 \,.$$
Now, you should be able to find the equation of the plane $P_1\,.$
The direction vector of the line is the normal vector to the plane which is $<4,4,-5>$
Thus you get $$x=4t, y=4t,z=-5t$$ for parametric and $$x/4=y/4=z/{-5}$$ for canonical form of the line.
Best Answer
Well, that line is given by $\langle x,y,z\rangle=\langle2t,-3t,-6t\rangle.$ We know the origin is a point on this line. What we need, then, is to find $t_0$ such that $2(2t_0)-3(-3t_0)-6(-6t_0)=-4$ (why?), then find the distance from the point $\langle2t_0,-3t_0,-6t_0\rangle$ to the origin.