If the straight line `(x-x_0)/l=(y-y_0)/m=(z-z_0)/n` is in the plane `ax+by+c+d=0` then which one of the following is correct?

Options

A. `l+m+n=0`
B. `a+b+c=0`
C. `a/1+b/m+c/n=0`
D. `al+bm+cn=0`

Answer

To solve the problem, we need to establish the relationship between the given straight line and the plane it lies in. Let's break down the steps:

### Step 1: Understand the Equation of the Line
The equation of the straight line is given in the symmetric form:
\[ \frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n} \]
This can be rewritten in vector form as:
\[ \mathbf{r} = \mathbf{r_0} + t \mathbf{d} \]
where \(\mathbf{r_0} = (x_0, y_0, z_0)\) is a point on the line, \(\mathbf{d} = (l, m, n)\) is the direction vector of the line, and \(t\) is a parameter.

### Step 2: Understand the Equation of the Plane
The equation of the plane is given as:
\[ ax + by + cz + d = 0 \]
The normal vector \(\mathbf{n}\) to the plane can be represented as:
\[ \mathbf{n} = (a, b, c) \]

### Step 3: Condition for the Line to Lie in the Plane
For the line to lie in the plane, the direction vector of the line must be perpendicular to the normal vector of the plane. This is expressed mathematically by the dot product:
\[ \mathbf{n} \cdot \mathbf{d} = 0 \]
This means:
\[ (a, b, c) \cdot (l, m, n) = 0 \]
Expanding this dot product gives:
\[ al + bm + cn = 0 \]

### Step 4: Conclusion
Thus, the correct condition that must hold for the line to lie in the plane is:
\[ al + bm + cn = 0 \]

### Final Answer
The correct option is:
\[ al + bm + cn = 0 \]

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