Through the intersection of π_1: 4x-y+3z-1=0 and π_2…

Through the intersection of $\pi_1:\ 4x-y+3z-1=0$ and $\pi_2:\ x+5y-z+2=0$, find the plane that: $1^\circ$ passes through $A(a,b,c)$; $2^\circ$ is parallel to the $y$-axis; $3^\circ$ is parallel to the $xOy$ plane; $4^\circ$ is perpendicular to $2x-y+5z-3=0$.

Step 1: The pencil

\[ (4x-y+3z-1)+\lambda(x+5y-z+2)=0 \]

\[ (4+\lambda)x+(5\lambda-1)y+(3-\lambda)z+(2\lambda-1)=0 \]

Here $\lambda=\infty$ gives $\pi_2$ itself. Equivalently, use $\alpha\pi_1+\beta\pi_2=0$ with $(\alpha,\beta)\ne(0,0)$.

Step 2: $1^\circ$ Through $A(a,b,c)$

\[(a+5b-c+2)(4x-y+3z-1)=(4a-b+3c-1)(x+5y-z+2)\]

This single formula covers all cases:

• If $A\notin\pi_2$, it gives $\lambda=-\tfrac{4a-b+3c-1}{a+5b-c+2}$.
• If $A\in\pi_2$ but $A\notin\pi_1$, it reduces to $\pi_2$.
• If $A$ lies on the line itself, both sides vanish identically and every plane of the pencil works.

Step 3: $2^\circ$ Parallel to $Oy$

The $y$-coefficient must vanish: $5\lambda-1=0$, so $\lambda=\tfrac15$.

\[21x+14z-3=0\]

Step 4: $3^\circ$ Parallel to $xOy$

This would need $4+\lambda=0$ and $5\lambda-1=0$ at the same time, which is impossible. The member $\pi_2$ (normal $\{1,5,-1\}$) does not work either.

\[\text{no such plane}\]

Geometric reason: the line $\pi_1\cap\pi_2$ has direction $\vec n_1\times\vec n_2=\{-14,7,21\}$, which is not horizontal. A horizontal plane cannot contain it.

Step 5: $4^\circ$ Perpendicular to $2x-y+5z-3=0$

\[ 2(4+\lambda)-(5\lambda-1)+5(3-\lambda)=24-8\lambda=0\ \Longrightarrow\ \lambda=3 . \]

\[7x+14y+5=0\]

Check: $\{7,14,0\}\cdot\{2,-1,5\}=0$.