The equation of a plane passing through the line of intersection of the planes x+2y+3z = 2 and x - y + z = 3 and at a distance `2/sqrt 3` from the point (3, 1, -1) is

Options

A. (a)`5x -11y + z=17`
B. (b) `sqrt2 x + y=3 sqrt2 -1`
C. (c) `x + y+ z= sqrt3`
D. (d) `x - sqrt2 y=1-sqrt2`

Answer

To find the equation of the plane that passes through the line of intersection of the planes \(x + 2y + 3z = 2\) and \(x - y + z = 3\) and is at a distance of \(\frac{2}{\sqrt{3}}\) from the point \((3, 1, -1)\), we can follow these steps:

### Step 1: Write the general equation of the plane
The equation of a plane passing through the line of intersection of two planes can be expressed as:
\[ A_1x + B_1y + C_1z + D_1 + \lambda (A_2x + B_2y + C_2z + D_2) = 0 \]
where \((A_1, B_1, C_1, D_1)\) and \((A_2, B_2, C_2, D_2)\) are the coefficients of the two planes.

For our planes:
1. \(x + 2y + 3z - 2 = 0\) (Plane 1)
2. \(x - y + z - 3 = 0\) (Plane 2)

We can express the equation of the required plane as:
\[ (x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0 \]

### Step 2: Combine the equations
Expanding this gives:
\[ (1 + \lambda)x + (2 - \lambda)y + (3 + \lambda)z + (-2 - 3\lambda) = 0 \]
This can be rewritten as:
\[ (1 + \lambda)x + (2 - \lambda)y + (3 + \lambda)z + (-2 - 3\lambda) = 0 \]
Let \(A = 1 + \lambda\), \(B = 2 - \lambda\), \(C = 3 + \lambda\), and \(D = -2 - 3\lambda\).

### Step 3: Use the distance formula
The distance \(d\) from a point \((x_1, y_1, z_1)\) to the plane \(Ax + By + Cz + D = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting \((x_1, y_1, z_1) = (3, 1, -1)\) and \(d = \frac{2}{\sqrt{3}}\):
\[ \frac{|(1 + \lambda) \cdot 3 + (2 - \lambda) \cdot 1 + (3 + \lambda) \cdot (-1) + (-2 - 3\lambda)|}{\sqrt{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2}} = \frac{2}{\sqrt{3}} \]

### Step 4: Simplify the numerator
Calculating the numerator:
\[ |3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda| = | -2\lambda | \]
So, we have:
\[ \frac{|-2\lambda|}{\sqrt{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2}} = \frac{2}{\sqrt{3}} \]

### Step 5: Simplify the denominator
Calculating the denominator:
\[ \sqrt{(1 + \lambda)^2 + (2 - \lambda)^2 + (3 + \lambda)^2} = \sqrt{(1 + 2\lambda + \lambda^2) + (4 - 4\lambda + \lambda^2) + (9 + 6\lambda + \lambda^2)} = \sqrt{3\lambda^2 + 4\lambda + 14} \]

### Step 6: Set up the equation
Now, we equate:
\[ \frac{2|\lambda|}{\sqrt{3\lambda^2 + 4\lambda + 14}} = \frac{2}{\sqrt{3}} \]
Cross-multiplying gives:
\[ 2|\lambda|\sqrt{3} = 2\sqrt{3\lambda^2 + 4\lambda + 14} \]
Squaring both sides:
\[ 4\lambda^2 \cdot 3 = 4(3\lambda^2 + 4\lambda + 14) \]
This simplifies to:
\[ 12\lambda^2 = 12\lambda^2 + 16\lambda + 56 \]
Thus:
\[ 0 = 16\lambda + 56 \implies \lambda = -\frac{56}{16} = -\frac{7}{2} \]

### Step 7: Substitute \(\lambda\) back into the plane equation
Substituting \(\lambda = -\frac{7}{2}\) back into the equation of the plane:
\[ (1 - \frac{7}{2})x + (2 + \frac{7}{2})y + (3 - \frac{7}{2})z + (-2 + \frac{21}{2}) = 0 \]
This simplifies to:
\[ -\frac{5}{2}x + \frac{11}{2}y - \frac{1}{2}z + \frac{17}{2} = 0 \]
Multiplying through by 2 to eliminate fractions:
\[ -5x + 11y - z + 17 = 0 \]
Rearranging gives:
\[ 5x - 11y + z - 17 = 0 \]

### Final Answer
Thus, the equation of the plane is:
\[ 5x - 11y + z - 17 = 0 \]