A ball is thrown upwards from the plane surface of the ground. Suppose the plane surface from which the ball is thrown also 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 :

$\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$

$\vec{n} = 3\hat{i} + 2\hat{j} + 4\hat{k}$

$\vec{D} = 2\hat{i} + 3\hat{j} + \hat{k}$

(a, b, c) → point of line

$\vec{A}\hat{i} + \vec{B}\hat{j} + \vec{C}\hat{k}$ vector || line

so equation of line → $\frac{x-a}{A}=\frac{y-b}{B}=\frac{z-c}{C}$

$\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$