Write the equation of the plane through M_1(2,-1,3) which…
Write the equation of the plane through $M_1(2,-1,3)$ which: $1^\circ$ cuts equal intercepts on the coordinate axes; $2^\circ$ passes through the $x$-axis; $3^\circ$ passes through the origin and the point $(1,1,1)$.
Explanation of Method
Three planes through M₁(2, −1, 3), each satisfying a different extra condition.
Step 1: $1^\circ$ Equal intercepts
$\tfrac xa+\tfrac ya+\tfrac za=1$, i.e. $x+y+z=a$. Through $M_1$: $a=2-1+3=4$.
\[x+y+z-4=0\]
Planes through the origin have no proper intercepts (all zero), so they are excluded. In particular $x+y+z=0$ does not contain $M_1$.
Step 2: $2^\circ$ Through the $x$-axis
We need $D=0$ and $A=0$, so the plane has the form $By+Cz=0$. Through $M_1$: $-B+3C=0$.
\[3y+z=0\]
Step 3: $3^\circ$ Through $O$ and $P(1,1,1)$
The normal is perpendicular to $\overrightarrow{OP}$ and $\overrightarrow{OM_1}$: $\vec n=\{1,1,1\}\times\{2,-1,3\}=\{4,-1,-3\}$.
\[4x-y-3z=0\quad(\text{equivalently }-4x+y+3z=0)\]
Checks: $P$ gives $4-1-3=0$, and $M_1$ gives $8+1-9=0$.
Result
\[\boxed{\,x+y+z-4=0\,}\]
\[\boxed{\,3y+z=0\,}\]
\[\boxed{\,4x-y-3z=0\quad(\text{equivalently }-4x+y+3z=0)\,}\]