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 plane passing through the points A, B and C is:

3x + 2y + 4z = 11

$\vec{A}=\hat{i}+2 \hat{k}$

$\vec{B}=3 \hat{i}-\hat{j}+\hat{k}$

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

so  $\vec{C A}=\vec{A}-\vec{C}=-2 \hat{j}+\hat{k}$

$\vec{C B}=\vec{B}-\vec{C}=2 \hat{i}-3 \hat{j}$

so  $\vec{n}$ (perpendicular to plane)

$\vec{n} = \vec{CA} × \vec{CB}$

so  $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 1\\ 2 & -3 & 0 \end{vmatrix}$

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

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

let point passing $\vec{a} = \vec{A} = \hat{i}+2\hat{k}$

so  $(\vec{r} - \vec{a}) . \vec{n} = 0$

$\vec{r} . \vec{n} = \vec{a} . \vec{n}$

= $(x\hat{i}+y\hat{j}+z\hat{k})(3\hat{i}+2\hat{j}+4\hat{k}) = (\hat{i}+0\hat{j}+2\hat{k})(3\hat{i}+2\hat{j}+4\hat{k})$

⇒ 3x + 2y + 4z

= 3 + 8

= 11