The vertices of `DeltaABC` are A(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. The z-coordinate of H is :

Options

A. 1
B. `1//2`
C. `1//6`
D. `1//3`

Answer

To find the z-coordinate of the orthocenter \( H \) of triangle \( ABC \) with vertices \( A(2, 0, 0) \), \( B(0, 1, 0) \), and \( C(0, 0, 2) \), we will follow these steps:

### Step 1: Find the equation of the plane containing triangle \( ABC \).

The intercept form of the equation of a plane is given by:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
where \( a \), \( b \), and \( c \) are the x, y, and z-intercepts respectively.

For triangle \( ABC \):
- The x-intercept is 2 (from point \( A(2, 0, 0) \)),
- The y-intercept is 1 (from point \( B(0, 1, 0) \)),
- The z-intercept is 2 (from point \( C(0, 0, 2) \)).

Thus, the equation of the plane is:
\[ \frac{x}{2} + \frac{y}{1} + \frac{z}{2} = 1 \]

### Step 2: Simplify the equation of the plane.

Multiplying through by 2 to eliminate the denominators gives:
\[ x + 2y + z = 2 \]

### Step 3: Find the direction ratios of the altitudes.

1. **Altitude from \( A \) to \( BC \)**:
- The direction ratios of line \( BC \) can be found from points \( B(0, 1, 0) \) and \( C(0, 0, 2) \):
\[ BC: (0-0, 0-1, 2-0) = (0, -1, 2) \]
- The direction ratios of altitude \( AH \) will be of the form \( (H_x - 2, H_y - 0, H_z - 0) = (H_x - 2, H_y, H_z) \).

2. **Using the perpendicularity condition**:
- Since \( AH \) is perpendicular to \( BC \), we have:
\[ (H_x - 2) \cdot 0 + H_y \cdot (-1) + H_z \cdot 2 = 0 \]
- This simplifies to:
\[ -H_y + 2H_z = 0 \quad \Rightarrow \quad H_y = 2H_z \quad \text{(1)} \]

### Step 4: Find the direction ratios of the altitude from \( B \) to \( AC \).

1. **Altitude from \( B \) to \( AC \)**:
- The direction ratios of line \( AC \) can be found from points \( A(2, 0, 0) \) and \( C(0, 0, 2) \):
\[ AC: (0-2, 0-0, 2-0) = (-2, 0, 2) \]
- The direction ratios of altitude \( BH \) will be of the form \( (H_x - 0, H_y - 1, H_z - 0) = (H_x, H_y - 1, H_z) \).

2. **Using the perpendicularity condition**:
- Since \( BH \) is perpendicular to \( AC \), we have:
\[ H_x \cdot (-2) + (H_y - 1) \cdot 0 + H_z \cdot 2 = 0 \]
- This simplifies to:
\[ -2H_x + 2H_z = 0 \quad \Rightarrow \quad H_x = H_z \quad \text{(2)} \]

### Step 5: Substitute equations (1) and (2) into the plane equation.

Substituting \( H_y = 2H_z \) and \( H_x = H_z \) into the plane equation \( x + 2y + z = 2 \):
\[ H_z + 2(2H_z) + H_z = 2 \]
\[ H_z + 4H_z + H_z = 2 \quad \Rightarrow \quad 6H_z = 2 \quad \Rightarrow \quad H_z = \frac{1}{3} \]

### Conclusion

The z-coordinate of the orthocenter \( H \) is:
\[ \boxed{\frac{1}{3}} \]