Determine D so that the line x-y+z+1=0, 2x-3y-z+D=0…

Determine $D$ so that the line $x-y+z+1=0,\ 2x-3y-z+D=0$ intersects the $z$-axis.

Step 1: Solution

The point $(0,0,c)$ must satisfy both equations. The first gives $c+1=0$, so $c=-1$. The second gives $-c+D=0$, so $D=c$.

\[D=-1,\qquad\text{meeting point }(0,0,-1)\]

The line is not the $z$-axis itself. Its direction is $\{1,-1,1\}\times\{2,-3,-1\}=\{4,3,-1\}$, so it crosses the axis at a single point.