Let image of the line [x-1]/3=[y-3]/5=[z-4]/2 in the plane 2x-y+z+3=0 be L. A plane 7x+ By+Cz+D=0 is such that it contains the line L and perpendicular to the plane 2x-y+z+3=0 then find the value of B+3C+D

To solve the problem, we need to analyze the given line and the plane, and then find the coefficients of the new plane that contains the line and is perpendicular to the existing plane. Let's break this down step by step.

Understanding the Line and the Plane

The line is given in symmetric form as:

• (x - 1)/3 = (y - 3)/5 = (z - 4)/2

This can be rewritten in parametric form. Letting the common ratio equal to a parameter \( t \), we have:

• x = 1 + 3t
• y = 3 + 5t
• z = 4 + 2t

Next, the plane is defined by the equation:

• 2x - y + z + 3 = 0

The normal vector of this plane can be extracted from the coefficients of \( x \), \( y \), and \( z \), which gives us:

• Normal vector \( \mathbf{n_1} = (2, -1, 1) \)

Finding the New Plane

The new plane is given by the equation:

• 7x + By + Cz + D = 0

Since this plane must contain the line \( L \), any point on the line must satisfy the plane's equation. Additionally, the plane must be perpendicular to the first plane, meaning its normal vector must be orthogonal to \( \mathbf{n_1} \).

Condition for Perpendicularity

The normal vector of the new plane is \( \mathbf{n_2} = (7, B, C) \). For the planes to be perpendicular, the dot product of their normal vectors must equal zero:

Dot Product Condition:

• 2 * 7 + (-1) * B + 1 * C = 0

This simplifies to:

• 14 - B + C = 0

Condition for Containing the Line

Next, we need to ensure that the plane contains the line. We can substitute the parametric equations of the line into the plane's equation:

Substituting \( x = 1 + 3t \), \( y = 3 + 5t \), and \( z = 4 + 2t \) into \( 7x + By + Cz + D = 0 \):

Expanding this gives:

• 7(1 + 3t) + B(3 + 5t) + C(4 + 2t) + D = 0

Distributing the terms results in:

• 7 + 21t + 3B + 5Bt + 4C + 2Ct + D = 0

Grouping the constant and \( t \) terms, we have:

• (7 + 3B + 4C + D) + (21 + 5B + 2C)t = 0

For this equation to hold for all \( t \), both coefficients must equal zero:

• 7 + 3B + 4C + D = 0
• 21 + 5B + 2C = 0

Solving the System of Equations

Now we have a system of two equations:

• 1. 3B + 4C + D = -7
• 2. 5B + 2C = -21

From the second equation, we can express \( D \) in terms of \( B \) and \( C \). Let's solve for \( C \) in terms of \( B \):

From \( 5B + 2C = -21 \):

• 2C = -21 - 5B
• C = (-21 - 5B)/2

Now substitute \( C \) back into the first equation:

Substituting into \( 3B + 4C + D = -7 \):

• 3B + 4((-21 - 5B)/2) + D = -7

Multiplying through by 2 to eliminate the fraction:

• 6B - 84 - 10B + 2D = -14

Combining like terms gives:

• -4B + 2D - 84 = -14

Rearranging this results in:

• 2D = 4B + 70
• D = 2B + 35

Final Steps to Find B + 3C + D

Now we have expressions for \( C \) and \( D \) in terms of \( B \):

• C = (-21 - 5B)/2
• D = 2B + 35

Now we can find \( B + 3C + D \):

Substituting \( C \) and \( D \):

• B + 3((-21 - 5B)/2) + (2B + 35)

Combining these terms:

• B - 63/2 - (15/2)B + 2B + 35

Combining like terms gives:

• (1 - 15/2 + 2)B + (35 - 63/2)

This simplifies to:

• (-11/2)B + (70/2 - 63/2) = (-11/2)B + (7/2)
</