Find the plane through the line x+2y+3z-4=0, 3x+z-5=0 that…

Find the plane through the line $x+2y+3z-4=0,\ 3x+z-5=0$ that cuts equal intercepts on the axes $Oy$ and $Oz$.

Step 1: Pencil and intercepts

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

\[ b=\frac{4+5\lambda}{2},\qquad c=\frac{4+5\lambda}{3+\lambda}\quad(\lambda\ne-3). \]

Step 2: Equal intercepts

Setting $b=c\ne0$: $3+\lambda=2$, so $\lambda=-1$.

\[2x-2y-2z-1=0\qquad\Big(a=\tfrac12,\ b=c=-\tfrac12\Big)\]

Check with points of the line:

• $P_1\left(0,-\tfrac{11}2,5\right)$: $0+11-10-1=0$;
• $P_2\left(\tfrac53,\tfrac76,0\right)$: $\tfrac{10}3-\tfrac73-1=0$.

Step 3: Degenerate and excluded cases

• $\lambda=-\tfrac45$ gives $-7x+10y+11z=0$, through the origin. Both “intercepts” are $0$, so they are only trivially equal. Usually this plane is not accepted, because a plane through $O$ does not cut segments on the axes.
• $\lambda=-3$ gives $-8x+2y+11=0$, which is parallel to $Oz$ (no $z$-intercept).
• The member $3x+z-5=0$ ($\lambda=\infty$) is parallel to $Oy$.

Neither of the last two works.