The plane 3x-y+4z-12=0 cuts the axes at the vertices of…
The plane $3x-y+4z-12=0$ cuts the axes at the vertices of triangle $ABC$, with $C$ on $Oz$. Write the altitude from $C$ and find its length.
Explanation of Method
The altitude from the vertex on the z-axis of the triangle cut out by a plane.
Step 1: Vertices
\[ A(4,0,0),\qquad B(0,-12,0),\qquad C(0,0,3). \]
Step 2: Direction of the altitude
It lies in the plane ($\perp\vec n=\{3,-1,4\}$) and is perpendicular to $AB\parallel\{1,3,0\}$: $\vec h=\{1,3,0\}\times\{3,-1,4\}=\{12,-4,-10\}\parallel\{6,-2,-5\}$.
\[\frac{x}{6}=\frac{y}{-2}=\frac{z-3}{-5}\]
Step 3: Foot and length
$z=0$ gives $s=\tfrac35$, so $H\left(\tfrac{18}5,-\tfrac65,0\right)$. $H$ is on $AB$ at parameter $\tfrac1{10}$.
\[|CH|=\sqrt{\tfrac{324}{25}+\tfrac{36}{25}+9}=\frac{3\sqrt{65}}5\approx4.84\]
Check with the area: $\overrightarrow{CA}\times\overrightarrow{CB}=\{-36,12,-48\}$, of length $12\sqrt{26}$, and $|AB|=4\sqrt{10}$. So
\[ |CH|=\frac{12\sqrt{26}}{4\sqrt{10}}=3\sqrt{\tfrac{13}5}=\frac{3\sqrt{65}}5.\ \checkmark \]
Result
\[\boxed{\,\frac{x}{6}=\frac{y}{-2}=\frac{z-3}{-5}\,}\]
\[\boxed{\,|CH|=\sqrt{\tfrac{324}{25}+\tfrac{36}{25}+9}=\frac{3\sqrt{65}}5\approx4.84\,}\]