In the plane Ax+3y-6z+D=0 determine A and D so that it…

In the plane $Ax+3y-6z+D=0$ determine $A$ and $D$ so that it passes through the intersection of $2x+y-z+3=0$ and $x-3y+5=0$.

Step 1: The pencil

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

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

The member $\lambda=\infty$ (the plane $x-3y+5=0$) has no $z$-term, so it cannot be $Ax+3y-6z+D=0$.

Step 2: Proportionality

The required plane is $k$ times a member of the pencil. The $z$-coefficients give $k=6$. Then the $y$-coefficients give

\[ 6(1-3\lambda)=3\ \Longrightarrow\ \lambda=\tfrac16 . \]

Hence $A=6\left(2+\tfrac16\right)=13$ and $D=6\left(3+\tfrac56\right)=23$.

\[A=13,\quad D=23:\qquad 13x+3y-6z+23=0\]

Step 3: Check

The line is $x=3y-5,\ z=7y-7$, with direction $\{3,1,7\}$.

• $P(-2,1,0)$: $-26+3+0+23=0$.
• $(-5,0,-7)$: $-65+0+42+23=0$.