If `P(alpha,beta,lambda)` is a vertex of an equilateral triangle PQR where vertex Q and R are `(-1,0,1)` and `(1,0,-1)` respectively, then P can lie on the plane

Options

A. `x+y+z+6=0`
B. `2x+4y+3z+20=0`
C. `x-y+z+12=0`
D. `x+y+z+3sqrt(2)=0`

Explanation

To find the equation of the plane on which the vertex P of the equilateral triangle PQR lies, we will follow these steps:

### Step 1: Identify the coordinates of points Q and R
Given:
- Q = (-1, 0, 1)
- R = (1, 0, -1)

### Step 2: Find the midpoint M of segment QR
The midpoint M can be calculated using the midpoint formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
Substituting the coordinates of Q and R:
\[ M = \left( \frac{-1 + 1}{2}, \frac{0 + 0}{2}, \frac{1 - 1}{2} \right) = (0, 0, 0) \]

### Step 3: Calculate the length of QR using the distance formula
The distance formula in 3D is given by:
\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Applying this to points Q and R:
\[ QR = \sqrt{(1 - (-1))^2 + (0 - 0)^2 + (-1 - 1)^2} = \sqrt{(2)^2 + (0)^2 + (-2)^2} = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 4: Determine the length of PM
Since PQR is an equilateral triangle, all sides are equal, thus:
\[ PQ = PR = QR = 2\sqrt{2} \]
Let PM = h. In triangle PMQ, we can use the relationship involving the angles:
- The angle at P is 60 degrees, and thus the angle at M is 30 degrees because the median divides the angle in half.

Using the tangent function:
\[ \tan(60^\circ) = \frac{PM}{MQ} \]
Where MQ is half of QR:
\[ MQ = \frac{QR}{2} = \frac{2\sqrt{2}}{2} = \sqrt{2} \]
Thus:
\[ \sqrt{3} = \frac{h}{\sqrt{2}} \implies h = \sqrt{2} \cdot \sqrt{3} = \sqrt{6} \]

### Step 5: Determine the plane equation
The distance from point P to the plane containing points Q and R must be less than or equal to PM:
\[ \text{Distance} \leq \sqrt{6} \]
The general equation of a plane can be represented as:
\[ ax + by + cz + d = 0 \]
For the plane containing points Q and R, we can derive the equation based on the coordinates. We can find the normal vector using the cross product of vectors QP and RP.

### Step 6: Check the options
We need to find the equation of the plane such that the distance from point P to the plane is less than or equal to \(\sqrt{6}\). After checking the options, we find that:
\[ x + y + z + 3\sqrt{2} = 0 \]
is the correct equation since it satisfies the distance condition.

### Conclusion
Thus, the vertex P can lie on the plane described by the equation:
\[ x + y + z + 3\sqrt{2} = 0 \]

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