Find the equation of plane which passes through the point `(1, 2,0)` and which is perpendicular to the plane `x-y+z=3 and 2x+y-z+4=0`.

Answer

To find the equation of the plane that passes through the point (1, 2, 0) and is perpendicular to the given planes, we can follow these steps:

### Step 1: Identify the normal vectors of the given planes
The equations of the two planes are:
1. Plane 1: \( x - y + z = 3 \)
2. Plane 2: \( 2x + y - z + 4 = 0 \)

From these equations, we can extract the normal vectors:
- For Plane 1, the normal vector \( \mathbf{p} \) is given by the coefficients of \( x, y, z \):
\[ \mathbf{p} = (1, -1, 1) \]

- For Plane 2, the normal vector \( \mathbf{q} \) is:
\[ \mathbf{q} = (2, 1, -1) \]

### Step 2: Find the cross product of the normal vectors
To find the normal vector of the required plane, we need to compute the cross product \( \mathbf{p} \times \mathbf{q} \):
\[ \mathbf{p} \times \mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \mathbf{i} \left((-1)(-1) - (1)(1)\right) - \mathbf{j} \left((1)(-1) - (1)(2)\right) + \mathbf{k} \left((1)(1) - (-1)(2)\right) \]
\[ = \mathbf{i} (1 - 1) - \mathbf{j} (-1 - 2) + \mathbf{k} (1 + 2) \]
\[ = 0\mathbf{i} + 3\mathbf{j} + 3\mathbf{k} \]
Thus, the normal vector of the required plane is:
\[ \mathbf{n} = (0, 3, 3) \]

### Step 3: Write the equation of the plane
The general equation of a plane can be expressed as:
\[ n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( (n_1, n_2, n_3) \) is the normal vector.

Substituting \( (x_0, y_0, z_0) = (1, 2, 0) \) and \( (n_1, n_2, n_3) = (0, 3, 3) \):
\[ 0(x - 1) + 3(y - 2) + 3(z - 0) = 0 \]
This simplifies to:
\[ 3(y - 2) + 3z = 0 \]
\[ 3y - 6 + 3z = 0 \]
Dividing the entire equation by 3:
\[ y + z - 2 = 0 \]
Thus, the equation of the required plane is:
\[ y + z = 2 \]

### Final Answer
The equation of the plane is:
\[ y + z = 2 \]

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