Find the equation of normal to the curve `y(x-2)(x-3)-x+7=0` at that point at which the curve meets X-axis.

Answer

To find the equation of the normal to the curve \( y(x-2)(x-3) - x + 7 = 0 \) at the point where the curve meets the X-axis, we can follow these steps:

### Step 1: Find the point where the curve meets the X-axis
The curve meets the X-axis where \( y = 0 \). We substitute \( y = 0 \) into the equation:

\[ 0(x-2)(x-3) - x + 7 = 0 \]

This simplifies to:

\[ -x + 7 = 0 \]

Solving for \( x \):

\[ x = 7 \]

Thus, the point where the curve meets the X-axis is \( (7, 0) \).

### Step 2: Differentiate the curve to find the slope of the tangent
We need to differentiate the equation \( y(x-2)(x-3) - x + 7 = 0 \) with respect to \( x \).

Using the product rule, we differentiate:

\[ \frac{d}{dx}[y(x-2)(x-3)] - \frac{d}{dx}[x] + \frac{d}{dx}[7] = 0 \]

This gives us:

\[ \frac{dy}{dx}(x-2)(x-3) + y\left(\frac{d}{dx}[(x-2)(x-3)]\right) - 1 = 0 \]

Calculating \( \frac{d}{dx}[(x-2)(x-3)] \):

\[ \frac{d}{dx}[(x-2)(x-3)] = (x-3) + (x-2) = 2x - 5 \]

Substituting back, we have:

\[ \frac{dy}{dx}(x-2)(x-3) + y(2x - 5) - 1 = 0 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Rearranging the equation gives:

\[ \frac{dy}{dx}(x-2)(x-3) = 1 - y(2x - 5) \]

Thus,

\[ \frac{dy}{dx} = \frac{1 - y(2x - 5)}{(x-2)(x-3)} \]

### Step 4: Evaluate the slope of the tangent at the point \( (7, 0) \)
Substituting \( x = 7 \) and \( y = 0 \):

\[ \frac{dy}{dx} = \frac{1 - 0(2(7) - 5)}{(7-2)(7-3)} = \frac{1}{(5)(4)} = \frac{1}{20} \]

### Step 5: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:

\[ \text{slope of normal} = -\frac{1}{\frac{1}{20}} = -20 \]

### Step 6: Write the equation of the normal
Using the point-slope form of the equation of a line:

\[ y - y_1 = m(x - x_1) \]

Substituting \( (x_1, y_1) = (7, 0) \) and \( m = -20 \):

\[ y - 0 = -20(x - 7) \]

This simplifies to:

\[ y = -20x + 140 \]

Rearranging gives:

\[ 20x + y - 140 = 0 \]

### Final Answer
The equation of the normal to the curve at the point where it meets the X-axis is:

\[ 20x + y - 140 = 0 \]