To solve the problem, we need to find the locus \( M \) of the feet of the perpendiculars drawn from points on the line \( L \) to the plane \( P_1 \). The steps are as follows:
### Step 1: Identify the equations of the planes
The equations of the planes are given as:
- Plane \( P_1: x + 2y - z + 1 = 0 \)
- Plane \( P_2: 2x - y + z - 1 = 0 \)
### Step 2: Find the direction ratios of the line \( L \)
The line \( L \) is parallel to the line of intersection of the two planes. To find the direction ratios of \( L \), we need to solve the system of equations formed by the normal vectors of the planes.
The normal vector of \( P_1 \) is \( (1, 2, -1) \) and for \( P_2 \) it is \( (2, -1, 1) \). The direction ratios of the line of intersection can be found using the cross product of these two normal vectors.
\[
\text{Direction Ratios} = (1, 2, -1) \times (2, -1, 1)
\]
Calculating the cross product:
\[
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & -1 \\
2 & -1 & 1
\end{vmatrix}
= \hat{i}(2 \cdot 1 - (-1) \cdot (-1)) - \hat{j}(1 \cdot 1 - (-1) \cdot 2) + \hat{k}(1 \cdot (-1) - 2 \cdot 2)
\]
\[
= \hat{i}(2 - 1) - \hat{j}(1 + 2) + \hat{k}(-1 - 4)
\]
\[
= \hat{i}(1) - \hat{j}(3) - \hat{k}(5)
\]
Thus, the direction ratios are \( (1, -3, -5) \).
### Step 3: Write the parametric equations of the line \( L \)
Since the line \( L \) passes through the origin, its parametric equations can be written as:
\[
\frac{x}{1} = \frac{y}{-3} = \frac{z}{-5} = t
\]
This gives:
\[
x = t, \quad y = -3t, \quad z = -5t
\]
### Step 4: Find the foot of the perpendicular from a point on line \( L \) to plane \( P_1 \)
The foot of the perpendicular from a point \( (x_0, y_0, z_0) \) to the plane \( ax + by + cz + d = 0 \) can be found using the formula:
\[
\left( x_0 - \frac{a}{a^2 + b^2 + c^2} (ax_0 + by_0 + cz_0 + d), y_0 - \frac{b}{a^2 + b^2 + c^2} (ax_0 + by_0 + cz_0 + d), z_0 - \frac{c}{a^2 + b^2 + c^2} (ax_0 + by_0 + cz_0 + d) \right)
\]
For plane \( P_1: x + 2y - z + 1 = 0 \) (where \( a = 1, b = 2, c = -1, d = 1 \)):
Substituting \( (x_0, y_0, z_0) = (t, -3t, -5t) \):
\[
ax_0 + by_0 + cz_0 + d = t + 2(-3t) - (-5t) + 1 = t - 6t + 5t + 1 = 0 + 1 = 1
\]
Now substituting into the foot of the perpendicular formula:
\[
\left( t - \frac{1}{6}, -3t - \frac{2}{6}, -5t + \frac{1}{6} \right) = \left( t - \frac{1}{6}, -3t - \frac{1}{3}, -5t + \frac{1}{6} \right)
\]
### Step 5: Find the locus \( M \)
The locus \( M \) can be expressed in terms of \( t \):
\[
M: \left( x + \frac{1}{6}, \frac{y + \frac{1}{3}}{-3}, \frac{z - \frac{1}{6}}{-5} \right)
\]
### Step 6: Check which points lie on \( M \)
Now we will check the given options to see if they satisfy the locus equation derived above.
1. **Option (a)**: \( (0, -\frac{5}{6}, -\frac{2}{3}) \)
- Check if it satisfies \( x + \frac{1}{6} = 0 \), \( y + \frac{1}{3} = -3(-\frac{5}{6}) \), \( z - \frac{1}{6} = -5(-\frac{2}{3}) \).
2. **Option (b)**: \( (-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}) \)
- Check if it satisfies the same conditions.
3. **Option (c)**: \( (-\frac{5}{6}, 0, \frac{1}{6}) \)
- Check if it satisfies the same conditions.
4. **Option (d)**: \( (-\frac{1}{3}, 0, \frac{2}{3}) \)
- Check if it satisfies the same conditions.
After checking, we find that options (a) and (b) satisfy the conditions of the locus \( M \).
### Final Answer
The points that lie on the locus \( M \) are:
- (a) \( (0, -\frac{5}{6}, -\frac{2}{3}) \)
- (b) \( (-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}) \)
