Find the equation of plane passing through the point `(1, 2, 3)` and having the vector `r=2hat(i)-hat(j)+3hat(k)` normal to it.

Answer

To find the equation of the plane passing through the point \( (1, 2, 3) \) and having the vector \( \mathbf{r} = 2\hat{i} - \hat{j} + 3\hat{k} \) as normal to it, we can follow these steps:

### Step 1: Identify the normal vector and the point
The normal vector \( \mathbf{n} \) to the plane is given as:
\[ \mathbf{n} = 2\hat{i} - \hat{j} + 3\hat{k} \]
The point through which the plane passes is:
\[ P(1, 2, 3) \]

### Step 2: Write the equation of the plane
The general equation of a plane in 3D space can be expressed as:
\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( (n_x, n_y, n_z) \) are the components of the normal vector.

### Step 3: Substitute the values
Substituting the values into the equation:
- \( n_x = 2 \)
- \( n_y = -1 \)
- \( n_z = 3 \)
- \( (x_0, y_0, z_0) = (1, 2, 3) \)

The equation becomes:
\[ 2(x - 1) - 1(y - 2) + 3(z - 3) = 0 \]

### Step 4: Expand the equation
Expanding this equation gives:
\[ 2x - 2 - y + 2 + 3z - 9 = 0 \]

### Step 5: Simplify the equation
Combining like terms:
\[ 2x - y + 3z - 9 = 0 \]
or rearranging gives:
\[ 2x - y + 3z = 9 \]

### Final Equation
Thus, the equation of the plane is:
\[ 2x - y + 3z = 9 \]

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