The vertices of `DeltaABC` are (2, 0, 0), B(0, 1, 0), C(0, 0, 2). Its orthocentre is H and circumcentre is S. P is a point equidistant from A, B, C and the origin O.
Q. PA is equal to :

Options

A. 1
B. `sqrt(2)`
C. `sqrt((3)/(2))`
D. `(3)/(2)`

Answer

To solve the problem, we need to find the distance \( PA \) where \( P \) is a point equidistant from the vertices \( A(2, 0, 0) \), \( B(0, 1, 0) \), \( C(0, 0, 2) \), and the origin \( O(0, 0, 0) \).

### Step-by-Step Solution:

**Step 1: Find the coordinates of the centroid \( G \) of triangle \( ABC \).**

The coordinates of the centroid \( G \) of a triangle with vertices \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) are given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right) \]

Substituting the coordinates of \( A(2, 0, 0) \), \( B(0, 1, 0) \), and \( C(0, 0, 2) \):
\[ G\left(\frac{2 + 0 + 0}{3}, \frac{0 + 1 + 0}{3}, \frac{0 + 0 + 2}{3}\right) = G\left(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right) \]

**Step 2: Write the equation of the line passing through the origin \( O \) and the centroid \( G \).**

The equation of the line can be expressed in parametric form as:
\[ \frac{x}{\frac{2}{3}} = \frac{y}{\frac{1}{3}} = \frac{z}{\frac{2}{3}} = k \]

From this, we can express \( x, y, z \) in terms of \( k \):
\[ x = \frac{2k}{3}, \quad y = \frac{k}{3}, \quad z = \frac{2k}{3} \]

**Step 3: Set up the condition for point \( P \) being equidistant from \( A, B, C, \) and \( O \).**

The distance from \( O \) to \( P \) is given by:
\[ OP = \sqrt{\left(\frac{2k}{3}\right)^2 + \left(\frac{k}{3}\right)^2 + \left(\frac{2k}{3}\right)^2} \]

The distance from \( O \) to \( A \) is:
\[ OA = \sqrt{(2 - 0)^2 + (0 - 0)^2 + (0 - 0)^2} = 2 \]

Since \( OP = PA \), we can express \( PA \) as:
\[ PA = \sqrt{\left(2 - \frac{2k}{3}\right)^2 + \left(0 - \frac{k}{3}\right)^2 + \left(0 - \frac{2k}{3}\right)^2} \]

**Step 4: Equate the distances and simplify.**

Setting \( OP = PA \):
\[ \sqrt{\left(\frac{2k}{3}\right)^2 + \left(\frac{k}{3}\right)^2 + \left(\frac{2k}{3}\right)^2} = \sqrt{\left(2 - \frac{2k}{3}\right)^2 + \left(-\frac{k}{3}\right)^2 + \left(-\frac{2k}{3}\right)^2} \]

Squaring both sides, we get:
\[ \frac{4k^2}{9} + \frac{k^2}{9} + \frac{4k^2}{9} = \left(2 - \frac{2k}{3}\right)^2 + \left(-\frac{k}{3}\right)^2 + \left(-\frac{2k}{3}\right)^2 \]

This simplifies to:
\[ \frac{9k^2}{9} = 4 - \frac{8k}{3} + \frac{4k^2}{9} + \frac{k^2}{9} + \frac{4k^2}{9} \]

**Step 5: Solve for \( k \).**

After simplification, we find:
\[ k^2 = 4 - \frac{8k}{3} + \frac{9k^2}{9} \]
This leads to a quadratic equation in \( k \). Solving this gives \( k = \frac{3}{2} \).

**Step 6: Find the coordinates of point \( P \).**

Substituting \( k = \frac{3}{2} \) into the parametric equations:
\[ P\left(\frac{2 \cdot \frac{3}{2}}{3}, \frac{\frac{3}{2}}{3}, \frac{2 \cdot \frac{3}{2}}{3}\right) = P(1, 0.5, 1) \]

**Step 7: Calculate \( PA \).**

Using the distance formula:
\[ PA = \sqrt{(2 - 1)^2 + (0 - 0.5)^2 + (0 - 1)^2} = \sqrt{1 + 0.25 + 1} = \sqrt{2.25} = \frac{3}{2} \]

Thus, the final answer is:
\[ \boxed{\frac{3}{2}} \]