Let a point P where position vector is $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ is called Lattice point if $x, y, z \in N$ . If atleast two of $x, y, z$ are equal then this Lattice point is called isosceles Lattice point. If all $x, y, z$ are equal then this Lattice point is called equilateral Lattice point.Which of the following statements is/are TRUE?

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Area of triangle formed by the isosceles Lattice points lying on the plane $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 4$ is $2 \sqrt{2}$ sq. units.

If a Lattice point is selected at random from Lattice points which satisfy $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) \leq 11,$ then the probability that the selected Lattice point is equilateral given that it is isosceles Lattice point is $\frac{1}{23}$ .

Area of triangle formed by the isosceles Lattice points lying on the plane $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 4$ is $\frac{\sqrt{3}}{2}$ sq.units.

answer is A, C.

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Detailed Solution

$x + y + z \leq 11$
For equilateral Lattice points $x = y = z = 1, 2, 3$
So three cases are possible for isosceles Lattice point $x = y \neq z$
$\begin{matrix} 1 & 1 & 9 \\ ⋮ \\ 5 & 5 & 1 \end{matrix}}$ 15 cases $\begin{matrix} 1 & 1 & 8 \\ ⋮ & ⋮ \\ 4 & 4 & 2 \end{matrix}}$ 12 cases
$\begin{matrix} 1 & 1 & 7 \\ 2 & 2 & 5 \\ 4 & 4 & 1 \end{matrix}}$ 9 cases $\begin{matrix} 1 & 1 & 6 \\ 2 & 2 & 4 \\ 3 & 3 & 2 \end{matrix}}$ 9 cases
$\begin{matrix} 1 & 1 & 5 \\ 2 & 2 & 3 \\ 3 & 3 & 1 \end{matrix}}$ 9 cases $\begin{matrix} 1 & 1 & 4 \end{matrix}}$ 3 cases
$\begin{matrix} 1 & 1 & 3 \\ 2 & 2 & 1 \end{matrix}}$ 6 cases $\begin{matrix} 1 & 1 & 2 \end{matrix}}$ 3 cases
Total number of isosceles Lattice points which are root equilateral points.
= 15 + 12 + 9 + 9 + 9 + 3 + 6 + 3 = 66.
Total number of isosceles Lattice point = 66 + 3 = 69.
$P = \frac{3}{69} = \frac{1}{23}$
Option C:
$x + y + z = 4$
$\begin{array}{l} x = y \neq z \\ 1 1 2 \end{array}}$ 3 cases
Points are $(1, 1, 2), (1, 2, 1), (2, 1, 1)$
Area of triangle is $\frac{\sqrt{3}}{2}$ sq.units.