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 coordinates of the foot of the perpendicular drawn from the maximum height of the ball to the ground are:

Plane through points $A,B,C$ $3x+2y+4z=11$ Normal vector of plane $\vec{n}=(3,2,4)$ Point $D(2,3,1)$ Foot of perpendicular formula $\left(x_1-\frac{a(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2},\; y_1-\frac{b(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2},\; z_1-\frac{c(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}\right)$ Plane $3x+2y+4z-11=0$ Substitute point $D$ $3(2)+2(3)+4(1)-11$ $=6+6+4-11$ $=5$ Denominator $3^2+2^2+4^2$ $=9+4+16$ $=29$ Coordinates of foot $x=2-\frac{3(5)}{29}=\frac{43}{29}$ $y=3-\frac{2(5)}{29}=\frac{77}{29}$ $z=1-\frac{4(5)}{29}=\frac{9}{29}$ Foot of perpendicular $\left(\frac{43}{29},\frac{77}{29},\frac{9}{29}\right)$