To solve the problem step by step, we will follow the outlined approach in the video transcript.
### Step 1: Identify the normal vectors of the given planes
The equations of the planes are:
1. \( P_1: 2x - y + z = 6 \)
2. \( P_2: x + 2y - z = 4 \)
From these equations, we can extract the normal vectors:
- For plane \( P_1 \), the normal vector \( \vec{N}_1 = \langle 2, -1, 1 \rangle \).
- For plane \( P_2 \), the normal vector \( \vec{N}_2 = \langle 1, 2, -1 \rangle \).
### Step 2: Calculate the cross product of the normal vectors
To find a normal vector \( \vec{N} \) for the required plane that is perpendicular to both \( \vec{N}_1 \) and \( \vec{N}_2 \), we calculate the cross product \( \vec{N} = \vec{N}_1 \times \vec{N}_2 \).
Using the determinant method:
\[
\vec{N} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
1 & 2 & -1
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{N} = \hat{i} \begin{vmatrix}
-1 & 1 \\
2 & -1
\end{vmatrix} - \hat{j} \begin{vmatrix}
2 & 1 \\
1 & -1
\end{vmatrix} + \hat{k} \begin{vmatrix}
2 & -1 \\
1 & 2
\end{vmatrix}
\]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix}
-1 & 1 \\
2 & -1
\end{vmatrix} = (-1)(-1) - (1)(2) = 1 - 2 = -1 \)
2. \( \begin{vmatrix}
2 & 1 \\
1 & -1
\end{vmatrix} = (2)(-1) - (1)(1) = -2 - 1 = -3 \)
3. \( \begin{vmatrix}
2 & -1 \\
1 & 2
\end{vmatrix} = (2)(2) - (-1)(1) = 4 + 1 = 5 \)
Putting it all together:
\[
\vec{N} = -\hat{i}(-1) - \hat{j}(-3) + \hat{k}(5) = \hat{i} + 3\hat{j} + 5\hat{k}
\]
Thus, \( \vec{N} = \langle -1, 3, 5 \rangle \).
### Step 3: Write the equation of the plane
The required plane passes through the point \( (1, 1, 1) \) and has the normal vector \( \vec{N} \). The equation of a plane can be written as:
\[
\vec{N} \cdot (\vec{r} - \vec{r_0}) = 0
\]
Where \( \vec{r_0} = \langle 1, 1, 1 \rangle \) and \( \vec{r} = \langle x, y, z \rangle \).
Substituting:
\[
\langle -1, 3, 5 \rangle \cdot \langle x - 1, y - 1, z - 1 \rangle = 0
\]
Expanding this gives:
\[
-1(x - 1) + 3(y - 1) + 5(z - 1) = 0
\]
This simplifies to:
\[
-x + 1 + 3y - 3 + 5z - 5 = 0
\]
Rearranging gives:
\[
-x + 3y + 5z - 7 = 0 \quad \text{or} \quad x - 3y - 5z + 7 = 0
\]
### Step 4: Calculate the distance from the origin to the plane
The distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
For our plane \( x - 3y - 5z + 7 = 0 \):
- \( A = 1, B = -3, C = -5, D = 7 \)
- The point is the origin \( (0, 0, 0) \).
Substituting into the distance formula:
\[
D = \frac{|1(0) - 3(0) - 5(0) + 7|}{\sqrt{1^2 + (-3)^2 + (-5)^2}} = \frac{|7|}{\sqrt{1 + 9 + 25}} = \frac{7}{\sqrt{35}}
\]
### Final Result
The distance from the origin to the required plane is:
\[
D = \frac{7}{\sqrt{35}} = \frac{7 \sqrt{5}}{5}
\]
