A ball is thrown upwards from the plane surface of the ground.
Suppose the plane surface from which the ball is thrown consists of the points
$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.
The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.
Using this information answer the question.

The equation of the perpendicular line drawn from the maximum height of the ball to the ground is:
🎥 Video solution / Text Solution of this question is given below:
Ground plane passes through points
$A(1,0,2),\; B(3,-1,1),\; C(1,2,1)$
Direction vectors
$\vec{AB}=(2,-1,-1)$
$\vec{AC}=(0,2,-1)$
Normal vector of plane
$\vec{n}=\vec{AB}\times\vec{AC}$
$\vec{n}=
\begin{vmatrix}
\mathbf{i}&\mathbf{j}&\mathbf{k}\\
2&-1&-1\\
0&2&-1
\end{vmatrix}
$
$=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$
So direction ratios of perpendicular line
$(3,2,4)$
The perpendicular line passes through point
$D(2,3,1)$
Hence equation of line
$\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$