Statement 1: Lthe cartesian equation of the plane `vecr=(hati-hatj)+lamda(hati+hatj+hatk)+mu(hati-2hatj+3hatk)` is `5x-2y-3z=7`
Statement 2: The non parametric form of the plane `vecr=veca+lamdavecb+muvecc` is `[(vecr,vecb,vecc)]=[(veca,vecb,vecc)]`

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze both statements and verify their correctness step by step.

### Step 1: Understanding the given vector equation of the plane
The vector equation of the plane is given as:
\[ \vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k}) \]
Here, we can identify:
- Point \( \vec{a} = \hat{i} - \hat{j} \)
- Direction vectors \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} - 2\hat{j} + 3\hat{k} \)

### Step 2: Finding the normal vector of the plane
The normal vector \( \vec{n} \) to the plane can be found using the cross product of the direction vectors \( \vec{b} \) and \( \vec{c} \):
\[ \vec{n} = \vec{b} \times \vec{c} \]

Calculating the cross product:
\[ \vec{b} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad \vec{c} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \]
Using the determinant formula for the cross product:
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot 3 - 1 \cdot (-2)) - \hat{j} (1 \cdot 3 - 1 \cdot 1) + \hat{k} (1 \cdot (-2) - 1 \cdot 1) \]
\[ = \hat{i} (3 + 2) - \hat{j} (3 - 1) + \hat{k} (-2 - 1) \]
\[ = 5\hat{i} - 2\hat{j} - 3\hat{k} \]

### Step 3: Writing the Cartesian equation of the plane
The Cartesian equation of the plane can be expressed as:
\[ \vec{n} \cdot (\vec{r} - \vec{a}) = 0 \]
Substituting \( \vec{n} = 5\hat{i} - 2\hat{j} - 3\hat{k} \) and \( \vec{a} = \hat{i} - \hat{j} \):
\[ 5(x - 1) - 2(y + 1) - 3(z - 0) = 0 \]
Expanding this:
\[ 5x - 5 - 2y - 2 - 3z = 0 \]
Rearranging gives:
\[ 5x - 2y - 3z = 7 \]

### Conclusion for Statement 1
Thus, Statement 1 is true: the Cartesian equation of the plane is indeed \( 5x - 2y - 3z = 7 \).

### Step 4: Understanding Statement 2
Statement 2 claims that the non-parametric form of the plane can be expressed as:
\[ [(\vec{r}, \vec{b}, \vec{c})] = [(\vec{a}, \vec{b}, \vec{c})] \]
This means that the vector \( \vec{r} \) lies in the plane formed by the vectors \( \vec{b} \) and \( \vec{c} \) and passes through point \( \vec{a} \).

### Conclusion for Statement 2
This is also true, as the non-parametric form of the plane is derived from the vector equation and confirms that \( \vec{r} \) lies in the plane defined by \( \vec{b} \) and \( \vec{c} \) through point \( \vec{a} \).

### Final Answer
Both statements are correct.

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