Find the equation of the plane which passes through the line of intersection of the planes `a_1x+b_1y+c_1z+d_1=0 and a_2x+b_2y+c_2z+d_2=0` and which is parallel to the line `(x-alpha)/l=(y-beta)/m=(z-gamma)/n`

Answer

To find the equation of the plane that passes through the line of intersection of the two given planes and is parallel to the specified line, we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the two planes are given as:
1. \( a_1x + b_1y + c_1z + d_1 = 0 \) (Equation 1)
2. \( a_2x + b_2y + c_2z + d_2 = 0 \) (Equation 2)

### Step 2: Formulate the equation of the required plane
The required plane can be expressed as a linear combination of the two given planes. Thus, we can write:
\[ \lambda_1 (a_1x + b_1y + c_1z + d_1) + \lambda_2 (a_2x + b_2y + c_2z + d_2) = 0 \]
where \(\lambda_1\) and \(\lambda_2\) are constants.

### Step 3: Simplify the equation
This can be rearranged to:
\[ (\lambda_1 a_1 + \lambda_2 a_2)x + (\lambda_1 b_1 + \lambda_2 b_2)y + (\lambda_1 c_1 + \lambda_2 c_2)z + (\lambda_1 d_1 + \lambda_2 d_2) = 0 \]

### Step 4: Identify the direction ratios of the line
The line is given in the symmetric form:
\[ \frac{x - \alpha}{l} = \frac{y - \beta}{m} = \frac{z - \gamma}{n} \]
The direction ratios of this line are \( (l, m, n) \).

### Step 5: Apply the condition for parallelism
For the plane to be parallel to the line, the normal vector of the plane must be perpendicular to the direction ratios of the line. This gives us the condition:
\[ (l, m, n) \cdot (\lambda_1 a_1 + \lambda_2 a_2, \lambda_1 b_1 + \lambda_2 b_2, \lambda_1 c_1 + \lambda_2 c_2) = 0 \]
This leads to:
\[ l(\lambda_1 a_1 + \lambda_2 a_2) + m(\lambda_1 b_1 + \lambda_2 b_2) + n(\lambda_1 c_1 + \lambda_2 c_2) = 0 \]

### Step 6: Solve for \(\lambda_1\) and \(\lambda_2\)
Rearranging gives:
\[ \lambda_1 (la_1 + mb_1 + nc_1) + \lambda_2 (la_2 + mb_2 + nc_2) = 0 \]
From this, we can express \(\lambda_2\) in terms of \(\lambda_1\):
\[ \lambda_2 = -\frac{la_1 + mb_1 + nc_1}{la_2 + mb_2 + nc_2} \lambda_1 \]

### Step 7: Substitute back into the plane equation
Substituting \(\lambda_2\) back into the equation of the plane gives us the final equation of the plane:
\[ \text{Final equation: } (la_2 + mb_2 + nc_2)(a_1x + b_1y + c_1z + d_1) - (la_1 + mb_1 + nc_1)(a_2x + b_2y + c_2z + d_2) = 0 \]

### Conclusion
Thus, the equation of the required plane is:
\[ (la_2 + mb_2 + nc_2)(a_1x + b_1y + c_1z + d_1) - (la_1 + mb_1 + nc_1)(a_2x + b_2y + c_2z + d_2) = 0 \]

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