A plane P passes through the line of intersection of the planes `2x-y+3z=2,  x+y-z=1` and the point `(1,0,1)`
If the plane P touches the sphere `x^2+y^2+z^2=r^2` then what is r equal to?

Options

A. A) `2/sqrt(29)` units
B. B) `4/sqrt(29)` units
C. C) `5/sqrt(29)` units
D. D) 1 unit

Answer

To solve the problem, we need to find the radius \( r \) of the sphere given that the plane \( P \) passes through the line of intersection of two planes and touches the sphere centered at the origin. We will follow these steps:

### Step 1: Identify the equations of the planes
The equations of the two planes are:
1. \( 2x - y + 3z = 2 \) (Plane 1)
2. \( x + y - z = 1 \) (Plane 2)

### Step 2: Find the equation of the plane \( P \)
The plane \( P \) can be expressed as a linear combination of the two planes. Thus, we can write:
\[ P: (2x - y + 3z - 2) + \lambda (x + y - z - 1) = 0 \]
where \( \lambda \) is a parameter.

### Step 3: Substitute the point (1, 0, 1) into the equation of the plane
We substitute \( x = 1 \), \( y = 0 \), and \( z = 1 \) into the equation of the plane \( P \):
\[ (2(1) - 0 + 3(1) - 2) + \lambda (1 + 0 - 1 - 1) = 0 \]
This simplifies to:
\[ (2 + 3 - 2) + \lambda (0 - 1) = 0 \]
\[ 3 - \lambda = 0 \]
Thus, we find:
\[ \lambda = 3 \]

### Step 4: Substitute \( \lambda \) back into the equation of the plane
Now, substituting \( \lambda = 3 \) back into the equation of the plane \( P \):
\[ P: (2x - y + 3z - 2) + 3(x + y - z - 1) = 0 \]
Expanding this gives:
\[ 2x - y + 3z - 2 + 3x + 3y - 3z - 3 = 0 \]
Combining like terms:
\[ (2x + 3x) + (-y + 3y) + (3z - 3z) - 5 = 0 \]
This simplifies to:
\[ 5x + 2y - 5 = 0 \]

### Step 5: Find the distance from the center of the sphere to the plane
The center of the sphere is at the origin \( (0, 0, 0) \). The distance \( d \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \) is given by:
\[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
For our plane \( 5x + 2y - 5 = 0 \), we have \( a = 5 \), \( b = 2 \), \( c = 0 \), and \( d = -5 \). Substituting \( (x_1, y_1, z_1) = (0, 0, 0) \):
\[ d = \frac{|5(0) + 2(0) - 5|}{\sqrt{5^2 + 2^2 + 0^2}} = \frac{|-5|}{\sqrt{25 + 4}} = \frac{5}{\sqrt{29}} \]

### Step 6: Conclusion
Since the plane touches the sphere, the distance \( d \) is equal to the radius \( r \):
\[ r = \frac{5}{\sqrt{29}} \]

### Final Answer
Thus, the radius \( r \) of the sphere is:
\[ \boxed{\frac{5}{\sqrt{29}}} \]