Consider the family of planes `x+y+z=c` where c is a parameter intersecting the coordinate axes P, Q andR and `alpha, beta and gamma` are the angles made by each member of this family with positive x, y and z-axes. Which of the following interpretations hold good got this family?

Options

A. Each member of this family is equally inclined with coordinate axes.
B. `sin^2(alpha)+sin^2(gamma)+sin^2(beta)=1`
C. `cos^2(alpha)+cos^2(beta)+cos^2(gamma)=2`
D. For c=3 area of the `trianglePQR is 3sqrt(3)` sq. units.

Answer

To solve the problem, we need to analyze the family of planes given by the equation \( x + y + z = c \), where \( c \) is a parameter. We will find the points where this plane intersects the coordinate axes and the angles it makes with these axes.

### Step 1: Find the intersection points with the coordinate axes

The plane \( x + y + z = c \) intersects the coordinate axes at points \( P \), \( Q \), and \( R \).

- **Intersection with the x-axis**: Set \( y = 0 \) and \( z = 0 \):
\[ x + 0 + 0 = c \implies x = c \implies P(c, 0, 0) \]

- **Intersection with the y-axis**: Set \( x = 0 \) and \( z = 0 \):
\[ 0 + y + 0 = c \implies y = c \implies Q(0, c, 0) \]

- **Intersection with the z-axis**: Set \( x = 0 \) and \( y = 0 \):
\[ 0 + 0 + z = c \implies z = c \implies R(0, 0, c) \]

### Step 2: Determine the angles \( \alpha \), \( \beta \), and \( \gamma \)

The angles made by the plane with the positive x, y, and z axes are denoted as \( \alpha \), \( \beta \), and \( \gamma \) respectively. The direction cosines of these angles can be given as follows:

- The direction cosines are:
\[ \cos \alpha = \frac{1}{\sqrt{3}}, \quad \cos \beta = \frac{1}{\sqrt{3}}, \quad \cos \gamma = \frac{1}{\sqrt{3}} \]

This is because the coefficients of \( x \), \( y \), and \( z \) in the equation \( x + y + z = c \) are all equal to 1.

### Step 3: Verify the relationships involving sine and cosine

Using the properties of direction cosines, we can verify the following relationships:

- **For sine**:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 \]

Calculating \( \sin \) values:
\[ \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left(\frac{1}{\sqrt{3}}\right)^2} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} \implies \sin^2 \alpha = \frac{2}{3} \]
Similarly,
\[ \sin^2 \beta = \frac{2}{3}, \quad \sin^2 \gamma = \frac{2}{3} \]
Thus,
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} = 2 \quad \text{(not equal to 1)} \]

- **For cosine**:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
Calculating:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \]

### Step 4: Area of triangle PQR

The area of triangle formed by points \( P \), \( Q \), and \( R \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
The lengths of the sides can be calculated as follows:
- Length of \( PQ \) = \( \sqrt{c^2 + c^2} = c\sqrt{2} \)
- Length of \( PR \) = \( \sqrt{c^2 + c^2} = c\sqrt{2} \)
- Length of \( QR \) = \( \sqrt{c^2 + c^2} = c\sqrt{2} \)

Using the formula for the area of a triangle formed by three points in 3D space, we can find the area when \( c = 3 \):
\[ \text{Area} = \frac{1}{2} \times c^2 \times \sqrt{3} = \frac{1}{2} \times 3^2 \times \sqrt{3} = \frac{9\sqrt{3}}{2} \]

### Conclusion

Based on the analysis:
1. Each member of this family is equally inclined with the coordinate axes.
2. The relationship \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 \) does not hold.
3. The relationship \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \) holds true.
4. The area of triangle \( PQR \) when \( c = 3 \) is \( \frac{9\sqrt{3}}{2} \).