Equation of any plane containing the line `(x-x_1)/(a)=(y-y_1)/(b)=(z-z_1)/(c)` is `A(x-x_1)+B(y-y_1)+C(z-z_1)=0` then pick correct alternatives

Options

A. `(A)/(a)=(B)/(b)=(C)/(c)` is true for the line to be perpendicular to the plane.
B. `A(a+3)+B(b-1)+C(c-2)=0`
C. `2aA+3bB+4cC=0`
D. `Aa+Bb+Cc=0`

Answer

To find the equation of any plane containing the line given by the parametric equations \((x - x_1)/a = (y - y_1)/b = (z - z_1)/c\), we can follow these steps:

### Step 1: Understand the given line equation
The line is represented in symmetric form as:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda \]
This implies:
\[ x = x_1 + a\lambda, \quad y = y_1 + b\lambda, \quad z = z_1 + c\lambda \]

### Step 2: Substitute the parametric equations into the plane equation
The general equation of a plane can be expressed as:
\[ A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 \]
Substituting the parametric equations into this plane equation, we get:
\[ A((x_1 + a\lambda) - x_1) + B((y_1 + b\lambda) - y_1) + C((z_1 + c\lambda) - z_1) = 0 \]
This simplifies to:
\[ A(a\lambda) + B(b\lambda) + C(c\lambda) = 0 \]

### Step 3: Factor out \(\lambda\)
We can factor out \(\lambda\) from the equation:
\[ \lambda(Aa + Bb + Cc) = 0 \]

### Step 4: Analyze the equation
For the equation \(\lambda(Aa + Bb + Cc) = 0\) to hold for all values of \(\lambda\), the term in the parentheses must equal zero:
\[ Aa + Bb + Cc = 0 \]

### Conclusion
Thus, the equation of the plane containing the line is given by:
\[ Aa + Bb + Cc = 0 \]