For the line `(x-1)/1=(y-2)/2=(z-3)/3,` which one of the following is correct? a. it lies in the plane `x-2y+z=0` b. it is same as line `x/1=y/2=z/3` c. it passes through `(2,3,5)` d. it is parallel t the plane`x-2y+z-6=0`

Options

A. It lies in the plane `x-2y+z=0`
B. It is same as line `(x)/(1)=(y)/(2)=(z)/(3)`
C. It passes through (2,3,5)
D. It is parallel to the plane `x-2y+z-6=0`

Explanation

To solve the problem step by step, let's analyze the given line and the options provided.

### Given Line:
The line is represented as:
\[ \frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3} \]
This can be rewritten in parametric form:
- \( x = 1 + t \)
- \( y = 2 + 2t \)
- \( z = 3 + 3t \)

Where \( t \) is a parameter.

### Step 1: Check if the line lies in the plane \( x - 2y + z = 0 \)

To check if the line lies in the plane, we need to see if the point through which the line passes satisfies the plane's equation. The line passes through the point \( (1, 2, 3) \).

Substituting \( x = 1 \), \( y = 2 \), and \( z = 3 \) into the plane equation:
\[ 1 - 2(2) + 3 = 1 - 4 + 3 = 0 \]
Since the left-hand side equals the right-hand side, the line **does lie in the plane**.

### Conclusion for Step 1:
**Option (a) is correct.**

### Step 2: Check if it is the same as the line \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \)

The line \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) can be represented in parametric form as:
- \( x = t \)
- \( y = 2t \)
- \( z = 3t \)

To check if these lines are the same, we can find a point on the second line. For \( t = 0 \), the point is \( (0, 0, 0) \).

Now we check if the point \( (0, 0, 0) \) satisfies the first line's equation:
\[ \frac{0-1}{1} = \frac{0-2}{2} = \frac{0-3}{3} \]
This simplifies to:
\[ -1 = -1 = -1 \]
Since both lines share the same direction ratios and the point \( (1, 2, 3) \) lies on both lines, they are indeed the same.

### Conclusion for Step 2:
**Option (b) is correct.**

### Step 3: Check if it passes through the point \( (2, 3, 5) \)

To check if the line passes through \( (2, 3, 5) \), we can set the parametric equations equal to these coordinates:
- \( 1 + t = 2 \) → \( t = 1 \)
- \( 2 + 2t = 3 \) → \( 2 + 2(1) = 4 \) (not equal)
- \( 3 + 3t = 5 \) → \( 3 + 3(1) = 6 \) (not equal)

Since the values do not match, the line does not pass through \( (2, 3, 5) \).

### Conclusion for Step 3:
**Option (c) is incorrect.**

### Step 4: Check if it is parallel to the plane \( x - 2y + z - 6 = 0 \)

To check if the line is parallel to the plane, we need to find the normal vector of the plane, which is given by the coefficients of \( x, y, z \) in the plane equation:
\[ \text{Normal vector} = (1, -2, 1) \]
The direction ratios of the line are \( (1, 2, 3) \).

To check for parallelism, we calculate the dot product:
\[ 1 \cdot 1 + 2 \cdot (-2) + 3 \cdot 1 = 1 - 4 + 3 = 0 \]
Since the dot product is zero, the line is perpendicular to the normal vector of the plane, which means it is parallel to the plane.

### Conclusion for Step 4:
**Option (d) is correct.**

### Final Summary:
- **Option (a)**: Correct
- **Option (b)**: Correct
- **Option (c)**: Incorrect
- **Option (d)**: Correct