Do the planes x+2y+3z=4 , 2x+y-z=3 and 3x+3y+2z=7 have…

Do the planes $x+2y+3z=4$, $2x+y-z=3$ and $3x+3y+2z=7$ have common points?

Step 1: Determinant

\[ \begin{vmatrix}1&2&3\\2&1&-1\\3&3&2\end{vmatrix}=5-14+9=0 . \]

There is no unique point, so we check consistency.

Step 2: Elimination

Row reduction gives $x+2y+3z=4$, $-3y-7z=-5$ and $0=0$. The ranks are equal (2), so the system is consistent. The third equation is simply the sum of the first two, because $1+2=3,\ 2+1=3,\ 3-1=2,\ 4+3=7$.

Step 3: Common line

With $z=t$: \[x=\frac{2+5t}{3},\quad y=\frac{5-7t}{3},\quad z=t\]

Check: the direction $\{5,-7,3\}$ is parallel to $\vec n_1\times\vec n_2=\{1,2,3\}\times\{2,1,-1\}=\{-5,7,-3\}$. The three planes belong to one pencil: they all pass through the line through $\left(\tfrac23,\tfrac53,0\right)$.