Determine the equation of the plane that passes through the point $P(2,3,-1)$ and contains the line of intersection of the planes $2x-y+3z=4$ and $x+2y-z=1$.

Step-by-step solution

1. Write the normals of the two given planes: ${\mathbf{n}}_1=(2,-1,3)$ and ${\mathbf{n}}_2=(1,2,-1)$.
2. Any plane containing their line of intersection has a normal that is a linear combination of ${\mathbf{n}}_1$ and ${\mathbf{n}}_2$: $\mathbf{n}=(2+\lambda ,-1+2\lambda ,3-\lambda )$.
3. The general equation of such a plane is $(2+\lambda )x+(-1+2\lambda )y+(3-\lambda )z+(-4-\lambda )=0$.
4. Substitute the coordinates of $P(2,3,-1)$ to determine $\lambda$:\n$(2+\lambda )(2)+(-1+2\lambda )(3)+(3-\lambda )(-1)+(-4-\lambda )=0$. \nSimplifying gives $8\lambda -6=0\Rightarrow \lambda =\frac{3}{4}$.
5. Insert $\lambda =\frac{3}{4}$ to get the normal vector $\mathbf{n}=(\frac{11}{4},\frac{1}{2},\frac{9}{4})$.
6. Multiply by 4 to clear fractions: normal becomes $(11,2,9)$.
7. The plane equation becomes $11x+2y+9z+C=0$.
8. Find $C$ by plugging $P(2,3,-1)$: $11\cdot 2+2\cdot 3+9\cdot (-1)+C=0\Rightarrow 22+6-9+C=0\Rightarrow C=-19$.
9. Thus the required plane equation is $11x+2y+9z-19=0$, or equivalently $11x+2y+9z=19$.

Concept tested: Equation of a plane through a point and a line of intersection

Published 15 September 2026 · How this solution was prepared and checked