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 find the distance \( PA \) where \( P \) is a point equidistant from points \( A \), \( B \), \( C \), and the origin \( O \), we will follow these steps:

### Step 1: Define the Coordinates
The vertices of triangle \( \Delta ABC \) are given as:
- \( A(2, 0, 0) \)
- \( B(0, 1, 0) \)
- \( C(0, 0, 2) \)
- The origin \( O(0, 0, 0) \)

Let the coordinates of point \( P \) be \( (x, y, z) \).

### Step 2: Set Up the Equations
Since \( P \) is equidistant from \( A \), \( B \), \( C \), and \( O \), we can write the following equations based on the distance formula:

1. \( PA = PB \)
2. \( PB = PC \)
3. \( PA = PO \)

Using the distance formula, we can express these distances as:

\[ PA = \sqrt{(x - 2)^2 + y^2 + z^2} \]
\[ PB = \sqrt{x^2 + (y - 1)^2 + z^2} \]
\[ PC = \sqrt{x^2 + y^2 + (z - 2)^2} \]
\[ PO = \sqrt{x^2 + y^2 + z^2} \]

### Step 3: Equate Distances
From \( PA = PB \):

\[ \sqrt{(x - 2)^2 + y^2 + z^2} = \sqrt{x^2 + (y - 1)^2 + z^2} \]

Squaring both sides gives:

\[ (x - 2)^2 + y^2 + z^2 = x^2 + (y - 1)^2 + z^2 \]

Cancelling \( z^2 \) from both sides:

\[ (x - 2)^2 + y^2 = x^2 + (y - 1)^2 \]

Expanding both sides:

\[ x^2 - 4x + 4 + y^2 = x^2 + y^2 - 2y + 1 \]

Cancelling \( x^2 \) and \( y^2 \):

\[ -4x + 4 = -2y + 1 \]

Rearranging gives:

\[ 4x - 2y = 3 \quad \text{(Equation 1)} \]

### Step 4: Equate Another Pair of Distances
From \( PB = PC \):

\[ \sqrt{x^2 + (y - 1)^2 + z^2} = \sqrt{x^2 + y^2 + (z - 2)^2} \]

Squaring both sides gives:

\[ x^2 + (y - 1)^2 + z^2 = x^2 + y^2 + (z - 2)^2 \]

Cancelling \( x^2 \):

\[ (y - 1)^2 + z^2 = y^2 + (z - 2)^2 \]

Expanding:

\[ y^2 - 2y + 1 + z^2 = y^2 + z^2 - 4z + 4 \]

Cancelling \( y^2 \) and \( z^2 \):

\[ -2y + 1 = -4z + 4 \]

Rearranging gives:

\[ 2y - 4z = 3 \quad \text{(Equation 2)} \]

### Step 5: Equate Distances to the Origin
From \( PA = PO \):

\[ \sqrt{(x - 2)^2 + y^2 + z^2} = \sqrt{x^2 + y^2 + z^2} \]

Squaring both sides gives:

\[ (x - 2)^2 + y^2 + z^2 = x^2 + y^2 + z^2 \]

Cancelling \( y^2 \) and \( z^2 \):

\[ (x - 2)^2 = x^2 \]

Expanding gives:

\[ x^2 - 4x + 4 = x^2 \]

Cancelling \( x^2 \):

\[ -4x + 4 = 0 \]

Thus, we find:

\[ x = 1 \]

### Step 6: Substitute \( x \) to Find \( y \) and \( z \)
Substituting \( x = 1 \) into Equation 1:

\[ 4(1) - 2y = 3 \implies 4 - 2y = 3 \implies 2y = 1 \implies y = \frac{1}{2} \]

Substituting \( x = 1 \) into Equation 2:

\[ 2\left(\frac{1}{2}\right) - 4z = 3 \implies 1 - 4z = 3 \implies -4z = 2 \implies z = -\frac{1}{2} \]

Thus, the coordinates of point \( P \) are \( (1, \frac{1}{2}, -\frac{1}{2}) \).

### Step 7: Calculate \( PA \)
Now we calculate \( PA \):

\[ PA = \sqrt{(1 - 2)^2 + \left(\frac{1}{2} - 0\right)^2 + \left(-\frac{1}{2} - 0\right)^2} \]
\[ = \sqrt{(-1)^2 + \left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \]
\[ = \sqrt{1 + \frac{1}{4} + \frac{1}{4}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \]

### Final Answer
Thus, the distance \( PA \) is \( \frac{\sqrt{6}}{2} \).