Let $P_1$ be the plane through the origin containing the vectors $[1,2,-1]$ and $[0,1,1]$. Let $P_2$ be the plane through the point $(1,1,1)$ parallel to the vectors $[-1,2,2]$ and $[3,4,-2]$
I know how to find the standard form of a plane that passes through a point and contains a line, but not one that contains two lines.
Find an equation for $P_1$ in normal form $ax+by+cz=0$
The planes $P_1$ and $P_2$ intersect in a line. Find a parametric equation for this line.
Best Answer
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 $P_1\,.$