If the equation of a plane P , passing through in the intersection of the planes, `x+4y-z+7=0 and 3x+y+5z=8` is `ax+by+6z=15` for some `a,b,c in R` , then the distance of the point (3,2,-1) form the plane P is ............

Answer

To solve the problem, we need to find the distance of the point (3, 2, -1) from the plane \( P \) defined by the equation \( ax + by + 6z = 15 \), where the plane \( P \) passes through the intersection of the planes \( P_1: x + 4y - z + 7 = 0 \) and \( P_2: 3x + y + 5z = 8 \).

### Step 1: Find the equation of the plane \( P \)

The equation of the plane \( P \) can be expressed as a linear combination of the equations of the planes \( P_1 \) and \( P_2 \):

\[ P: P_1 + \lambda P_2 = 0 \]

Substituting the equations of \( P_1 \) and \( P_2 \):

\[ (x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0 \]

Expanding this gives:

\[ x + 4y - z + 7 + 3\lambda x + \lambda y + 5\lambda z - 8\lambda = 0 \]

Combining like terms results in:

\[ (1 + 3\lambda)x + (4 + \lambda)y + (-1 + 5\lambda)z + (7 - 8\lambda) = 0 \]

### Step 2: Compare with the given plane equation

We know that the plane \( P \) can also be expressed as:

\[ ax + by + 6z - 15 = 0 \]

Setting the coefficients equal gives us the following system of equations:

1. \( a = 1 + 3\lambda \)
2. \( b = 4 + \lambda \)
3. \( 6 = -1 + 5\lambda \)
4. \( -15 = 7 - 8\lambda \)

### Step 3: Solve for \( \lambda \)

From the third equation:

\[ 6 = -1 + 5\lambda \implies 5\lambda = 7 \implies \lambda = \frac{7}{5} \]

### Step 4: Substitute \( \lambda \) back to find \( a \) and \( b \)

Substituting \( \lambda = \frac{7}{5} \) into the equations for \( a \) and \( b \):

1. \( a = 1 + 3\left(\frac{7}{5}\right) = 1 + \frac{21}{5} = \frac{26}{5} \)
2. \( b = 4 + \frac{7}{5} = \frac{20}{5} + \frac{7}{5} = \frac{27}{5} \)

### Step 5: Write the equation of the plane \( P \)

Now we can write the equation of the plane \( P \):

\[ \frac{26}{5}x + \frac{27}{5}y + 6z = 15 \]

Multiplying through by 5 to eliminate the fractions:

\[ 26x + 27y + 30z = 75 \]

### Step 6: Calculate the distance from the point (3, 2, -1)

The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \) is:

\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, \( A = 26 \), \( B = 27 \), \( C = 30 \), and \( D = -75 \). The point is \( (3, 2, -1) \).

Calculating \( D \):

\[ D = \frac{|26(3) + 27(2) + 30(-1) - 75|}{\sqrt{26^2 + 27^2 + 30^2}} \]

Calculating the numerator:

\[ = |78 + 54 - 30 - 75| = |27| = 27 \]

Calculating the denominator:

\[ = \sqrt{676 + 729 + 900} = \sqrt{2305} \]

Thus, the distance is:

\[ D = \frac{27}{\sqrt{2305}} \]

### Final Answer

The distance of the point (3, 2, -1) from the plane \( P \) is \( \frac{27}{\sqrt{2305}} \).