Do the planes 2x-4y+3z=1 , x-2y+4z=3 and 3x-y+5z=2 have a…

Do the planes $2x-4y+3z=1$, $x-2y+4z=3$ and $3x-y+5z=2$ have a common point?

Explanation of Method

Three planes with a non-zero determinant meet in exactly one point.

Step 1: Determinant

\[ \Delta=\begin{vmatrix}2&-4&3\\1&-2&4\\3&-1&5\end{vmatrix}=2(-10+4)-(-4)(5-12)+3(-1+6)=-12-28+15=-25\ne0 . \]

Hence there is exactly one common point.

Step 2: Solving

Subtracting twice the second equation from the first gives $-5z=-5$, so $z=1$. Then $3x-y=-3$ and $x-2y=-1$, which give $x=-1$ and $y=0$.

\[M(-1,\,0,\,1)\]

Check: $-2+0+3=1$, $-1+0+4=3$ and $-3-0+5=2$. $\checkmark$

Result

\[\boxed{\,M(-1,\,0,\,1)\,}\]

Remark

The source's first computation of $\Delta$ is correct ($-25$). The repeated “recheck” and “correction” lines in the source that follow it are unnecessary and can be ignored.