(a) Find the length and the foot of the perpendicular from :
P (1,1,2) to the plane 2x - 2y + 4z + 5 = 0
(b) Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x - 3y + 4z - 6 = 0.

Answer

To solve the given problems, we will follow a systematic approach for each part.

### Part (a): Find the length and the foot of the perpendicular from point P(1, 1, 2) to the plane 2x - 2y + 4z + 5 = 0.

**Step 1: Identify the plane and the point.**
- The point P is given as (1, 1, 2).
- The equation of the plane is 2x - 2y + 4z + 5 = 0.

**Step 2: Determine the direction ratios of the normal to the plane.**
- The coefficients of x, y, and z in the plane equation give us the direction ratios of the normal vector to the plane.
- Here, the direction ratios are (2, -2, 4).

**Step 3: Write the parametric equations of the line perpendicular to the plane.**
- The line passing through point P(1, 1, 2) in the direction of the normal vector can be expressed as:
\[ \frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z - 2}{4} = k \]
This gives us:
- \( x = 2k + 1 \)
- \( y = -2k + 1 \)
- \( z = 4k + 2 \)

**Step 4: Substitute the parametric equations into the plane equation.**
- Substitute \( x, y, z \) into the plane equation:
\[ 2(2k + 1) - 2(-2k + 1) + 4(4k + 2) + 5 = 0 \]
Simplifying this:
\[ 4k + 2 + 4k - 2 + 16k + 8 + 5 = 0 \]
\[ 24k + 13 = 0 \implies k = -\frac{13}{24} \]

**Step 5: Find the coordinates of the foot of the perpendicular.**
- Substitute \( k \) back into the parametric equations:
- \( x_1 = 2(-\frac{13}{24}) + 1 = \frac{12 - 13}{24} = -\frac{1}{24} \)
- \( y_1 = -2(-\frac{13}{24}) + 1 = \frac{26}{24} + 1 = \frac{50}{24} = \frac{25}{12} \)
- \( z_1 = 4(-\frac{13}{24}) + 2 = -\frac{52}{24} + 2 = -\frac{52}{24} + \frac{48}{24} = -\frac{4}{24} = -\frac{1}{6} \)

Thus, the foot of the perpendicular is \( (-\frac{1}{24}, \frac{25}{12}, -\frac{1}{6}) \).

**Step 6: Calculate the length of the perpendicular.**
- The formula for the length of the perpendicular from point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \) is:
\[ \text{Length} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
- Here, \( A = 2, B = -2, C = 4, D = 5 \):
\[ \text{Length} = \frac{|2(-\frac{1}{24}) - 2(\frac{25}{12}) + 4(-\frac{1}{6}) + 5|}{\sqrt{2^2 + (-2)^2 + 4^2}} \]
Simplifying the numerator:
\[ = \frac{|-\frac{2}{24} - \frac{50}{12} - \frac{4}{6} + 5|}{\sqrt{4 + 4 + 16}} = \frac{|- \frac{1}{12} - \frac{50}{12} - \frac{8}{12} + \frac{60}{12}|}{\sqrt{24}} = \frac{|-\frac{59}{12} + \frac{60}{12}|}{\sqrt{24}} = \frac{| \frac{1}{12}|}{\sqrt{24}} = \frac{1}{12\sqrt{24}} = \frac{1}{12 \cdot 2\sqrt{6}} = \frac{1}{24\sqrt{6}} \]

### Part (b): Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x - 3y + 4z - 6 = 0.

**Step 1: Identify the plane and the point.**
- The origin is given as (0, 0, 0).
- The equation of the plane is 2x - 3y + 4z - 6 = 0.

**Step 2: Determine the direction ratios of the normal to the plane.**
- The direction ratios are (2, -3, 4).

**Step 3: Write the parametric equations of the line perpendicular to the plane.**
- The line passing through the origin in the direction of the normal vector can be expressed as:
\[ \frac{x}{2} = \frac{y}{-3} = \frac{z}{4} = k \]
This gives us:
- \( x = 2k \)
- \( y = -3k \)
- \( z = 4k \)

**Step 4: Substitute the parametric equations into the plane equation.**
- Substitute \( x, y, z \) into the plane equation:
\[ 2(2k) - 3(-3k) + 4(4k) - 6 = 0 \]
Simplifying this:
\[ 4k + 9k + 16k - 6 = 0 \implies 29k - 6 = 0 \implies k = \frac{6}{29} \]

**Step 5: Find the coordinates of the foot of the perpendicular.**
- Substitute \( k \) back into the parametric equations:
- \( x_1 = 2(\frac{6}{29}) = \frac{12}{29} \)
- \( y_1 = -3(\frac{6}{29}) = -\frac{18}{29} \)
- \( z_1 = 4(\frac{6}{29}) = \frac{24}{29} \)

Thus, the foot of the perpendicular from the origin to the plane is \( \left(\frac{12}{29}, -\frac{18}{29}, \frac{24}{29}\right) \).

### Summary of Results:
- (a) The foot of the perpendicular from point P(1, 1, 2) to the plane is \( (-\frac{1}{24}, \frac{25}{12}, -\frac{1}{6}) \) and the length of the perpendicular is \( \frac{1}{24\sqrt{6}} \).
- (b) The coordinates of the foot of the perpendicular from the origin to the plane are \( \left(\frac{12}{29}, -\frac{18}{29}, \frac{24}{29}\right) \).