Let L be a line obtained from the intersection of two planes x+2y+z = 6 and y+2z=4 . If point `P(alpha,beta,gamma)` is the foot of perpendicular from (3,2,1) on L then the value of `21(alpha+beta+gamma)` equals .

Options

A. 142
B. 68
C. 136
D. 102

Answer

To solve the problem, we need to find the coordinates of the foot of the perpendicular from the point \( (3, 2, 1) \) to the line \( L \) formed by the intersection of the two planes given by the equations:

1. \( x + 2y + z = 6 \)
2. \( y + 2z = 4 \)

### Step 1: Find the Direction Ratios of the Line \( L \)

The direction ratios of the line formed by the intersection of two planes can be found using the normal vectors of the planes.

- For the first plane \( x + 2y + z = 6 \), the normal vector \( \mathbf{n_1} = (1, 2, 1) \).
- For the second plane \( y + 2z = 4 \), we can rewrite it as \( 0x + 1y + 2z = 4 \), giving the normal vector \( \mathbf{n_2} = (0, 1, 2) \).

The direction ratios of the line \( L \) can be found using the cross product of the two normal vectors:

\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{d} = \mathbf{i}(2 \cdot 2 - 1 \cdot 1) - \mathbf{j}(1 \cdot 2 - 0 \cdot 1) + \mathbf{k}(1 \cdot 1 - 0 \cdot 2) \]
\[ = \mathbf{i}(4 - 1) - \mathbf{j}(2) + \mathbf{k}(1) \]
\[ = 3\mathbf{i} - 2\mathbf{j} + 1\mathbf{k} \]

Thus, the direction ratios of the line \( L \) are \( (3, -2, 1) \).

### Step 2: Find a Point on the Line \( L \)

To find a point on the line, we can solve the system of equations given by the two planes. We can set \( z = 0 \) and solve for \( x \) and \( y \):

From the second plane:
\[ y + 2(0) = 4 \implies y = 4 \]

Substituting \( y = 4 \) into the first plane:
\[ x + 2(4) + 0 = 6 \implies x + 8 = 6 \implies x = -2 \]

Thus, a point on the line \( L \) is \( (-2, 4, 0) \).

### Step 3: Parametric Equations of the Line \( L \)

Using the point \( (-2, 4, 0) \) and the direction ratios \( (3, -2, 1) \), we can write the parametric equations of the line:

\[ x = -2 + 3t, \quad y = 4 - 2t, \quad z = 0 + t \]

### Step 4: Find the Foot of the Perpendicular from Point \( (3, 2, 1) \)

Let \( P(\alpha, \beta, \gamma) \) be the foot of the perpendicular from the point \( (3, 2, 1) \) to the line \( L \). The coordinates of \( P \) can be expressed in terms of the parameter \( t \):

\[ P(t) = (-2 + 3t, 4 - 2t, t) \]

The vector \( AP \) from \( A(3, 2, 1) \) to \( P(t) \) is:

\[ AP = ( (-2 + 3t) - 3, (4 - 2t) - 2, t - 1 ) = (3t - 5, -2t + 2, t - 1) \]

The direction ratios of line \( L \) are \( (3, -2, 1) \). For \( AP \) to be perpendicular to \( L \), their dot product must equal zero:

\[ (3, -2, 1) \cdot (3t - 5, -2t + 2, t - 1) = 0 \]

Calculating the dot product:

\[ 3(3t - 5) - 2(-2t + 2) + 1(t - 1) = 0 \]
\[ 9t - 15 + 4t - 4 + t - 1 = 0 \]
\[ 14t - 20 = 0 \implies t = \frac{20}{14} = \frac{10}{7} \]

### Step 5: Find Coordinates of Point \( P \)

Substituting \( t = \frac{10}{7} \) into the parametric equations:

\[ \alpha = -2 + 3\left(\frac{10}{7}\right) = -2 + \frac{30}{7} = \frac{30 - 14}{7} = \frac{16}{7} \]
\[ \beta = 4 - 2\left(\frac{10}{7}\right) = 4 - \frac{20}{7} = \frac{28 - 20}{7} = \frac{8}{7} \]
\[ \gamma = \frac{10}{7} \]

### Step 6: Calculate \( 21(\alpha + \beta + \gamma) \)

Now, we calculate \( \alpha + \beta + \gamma \):

\[ \alpha + \beta + \gamma = \frac{16}{7} + \frac{8}{7} + \frac{10}{7} = \frac{34}{7} \]

Finally, we find \( 21(\alpha + \beta + \gamma) \):

\[ 21\left(\frac{34}{7}\right) = 3 \cdot 34 = 102 \]

Thus, the final answer is:

\[ \boxed{102} \]