let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes `P_1 : x + 2y-z +1 = 0` and `P_2 : 2x-y + z-1 = 0`, Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane `P_1`. Which of the following points lie(s) on M?

Options

A. `(0,-(5)/(6),-(2)/(3))`
B. `(-(1)/(6),-(1)/(3),(1)/(6))`
C. `(-(5)/(6),0,(1)/(6))`
D. `(-(1)/(3),0,(2)/(3))`

Explanation

To solve the problem, we need to find the locus of the feet of the perpendiculars drawn from points on the line \( L \) to the plane \( P_1 \). The line \( L \) passes through the origin and is at a constant distance from the two given planes \( P_1 \) and \( P_2 \).

### Step 1: Identify the normal vectors of the planes
The equations of the planes are:
- \( P_1: x + 2y - z + 1 = 0 \)
- \( P_2: 2x - y + z - 1 = 0 \)

The normal vector of plane \( P_1 \) is \( \mathbf{n_1} = (1, 2, -1) \) and for plane \( P_2 \) it is \( \mathbf{n_2} = (2, -1, 1) \).

### Step 2: Find the direction ratios of line \( L \)
Since the line \( L \) is at a constant distance from both planes, it must be parallel to the line of intersection of the two planes. To find the direction ratios of this line, we can take the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

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

Calculating the determinant, we get:
\[ \mathbf{d} = \mathbf{i} \left( 2 \cdot 1 - (-1) \cdot (-1) \right) - \mathbf{j} \left( 1 \cdot 1 - (-1) \cdot 2 \right) + \mathbf{k} \left( 1 \cdot (-1) - 2 \cdot 2 \right) \]
\[ = \mathbf{i} (2 - 1) - \mathbf{j} (1 + 2) + \mathbf{k} (-1 - 4) \]
\[ = \mathbf{i} (1) - \mathbf{j} (3) - \mathbf{k} (5) \]
Thus, the direction ratios of line \( L \) are \( (1, -3, -5) \).

### Step 3: Parametric equations of line \( L \)
The parametric equations of line \( L \) passing through the origin are:
\[ x = t, \quad y = -3t, \quad z = -5t \]

### Step 4: Find the foot of the perpendicular from a point on line \( L \) to plane \( P_1 \)
To find the foot of the perpendicular from a point \( (t, -3t, -5t) \) to the plane \( P_1 \), we use the formula for the foot of the perpendicular from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \):
\[ \text{Foot} = \left( x_0 - \frac{A(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2}, y_0 - \frac{B(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2}, z_0 - \frac{C(Ax_0 + By_0 + Cz_0 + D)}{A^2 + B^2 + C^2} \right) \]

For plane \( P_1 \):
- \( A = 1, B = 2, C = -1, D = 1 \)
- \( (x_0, y_0, z_0) = (t, -3t, -5t) \)

Calculating \( Ax_0 + By_0 + Cz_0 + D \):
\[ = 1(t) + 2(-3t) - 1(-5t) + 1 = t - 6t + 5t + 1 = 0 + 1 = 1 \]

Now, substituting into the foot of the perpendicular formula:
\[ \text{Foot} = \left( t - \frac{1(1)}{1^2 + 2^2 + (-1)^2}, -3t - \frac{2(1)}{1^2 + 2^2 + (-1)^2}, -5t - \frac{-1(1)}{1^2 + 2^2 + (-1)^2} \right) \]
\[ = \left( t - \frac{1}{6}, -3t - \frac{2}{6}, -5t + \frac{1}{6} \right) \]
\[ = \left( t - \frac{1}{6}, -3t - \frac{1}{3}, -5t + \frac{1}{6} \right) \]

### Step 5: Determine the locus \( M \)
As \( t \) varies, the coordinates of the foot of the perpendicular describe a locus \( M \).

### Final Step: Check which points lie on locus \( M \)
To find specific points that lie on \( M \), we can substitute values into the parametric equations derived above and check against the options given in the problem.