Find the equation of a plane passes through the point `(1,2,3)` and cuts equal intercepts on the co-ordinate axes.

Answer

To find the equation of a plane that passes through the point \( (1, 2, 3) \) and cuts equal intercepts on the coordinate axes, we can follow these steps:

### Step 1: Understand the intercept form of the equation of a plane
The equation of a plane in intercept form is given by:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
where \( a \), \( b \), and \( c \) are the x, y, and z-intercepts of the plane, respectively.

### Step 2: Set equal intercepts
Since the plane cuts equal intercepts on the coordinate axes, we have:
\[ a = b = c \]
Let’s denote this common intercept as \( a \). Therefore, the equation of the plane can be rewritten as:
\[ \frac{x}{a} + \frac{y}{a} + \frac{z}{a} = 1 \]

### Step 3: Simplify the equation
This simplifies to:
\[ \frac{x + y + z}{a} = 1 \]
Multiplying through by \( a \), we get:
\[ x + y + z = a \]

### Step 4: Substitute the point into the equation
Since the plane passes through the point \( (1, 2, 3) \), we can substitute these coordinates into the equation:
\[ 1 + 2 + 3 = a \]
Calculating the left side gives:
\[ 6 = a \]

### Step 5: Write the final equation of the plane
Now, substituting \( a = 6 \) back into the equation of the plane, we have:
\[ x + y + z = 6 \]

### Conclusion
Thus, the equation of the plane that passes through the point \( (1, 2, 3) \) and cuts equal intercepts on the coordinate axes is:
\[ \boxed{x + y + z = 6} \]