Find the equation fo the plane passing through the point (1,1,1) and containing the line :
`vec(r) = (-3 hati + hatj + 5 hatk) + lambda (3 hati - hatj + 5 hatk)`.
Also , show that the plane contains the line :
`vec(r) = (-hati + 2 hatj + 5 hatk ) + lambda (hati - 2 hatj - 5 hatk)`.

Answer

To find the equation of the plane passing through the point (1, 1, 1) and containing the line given by the vector equation:

\[ \vec{r} = (-3 \hat{i} + \hat{j} + 5 \hat{k}) + \lambda (3 \hat{i} - \hat{j} + 5 \hat{k}), \]

we will follow these steps:

### Step 1: Identify the point and direction vector of the line
The line can be expressed in terms of a point and a direction vector. The point on the line is:

\[ \vec{A} = -3 \hat{i} + \hat{j} + 5 \hat{k}, \]

and the direction vector of the line is:

\[ \vec{b} = 3 \hat{i} - \hat{j} + 5 \hat{k}. \]

### Step 2: Find the vector from the point on the plane to the point on the line
Let the point on the plane be:

\[ \vec{P} = 1 \hat{i} + 1 \hat{j} + 1 \hat{k}. \]

Now, we find the vector from point A to point P:

\[ \vec{AP} = \vec{P} - \vec{A} = (1 + 3) \hat{i} + (1 - 1) \hat{j} + (1 - 5) \hat{k} = 4 \hat{i} + 0 \hat{j} - 4 \hat{k}. \]

### Step 3: Find the normal vector to the plane
The normal vector \(\vec{n}\) to the plane can be found by taking the cross product of \(\vec{AP}\) and \(\vec{b}\):

\[ \vec{AP} = 4 \hat{i} + 0 \hat{j} - 4 \hat{k}, \]
\[ \vec{b} = 3 \hat{i} - \hat{j} + 5 \hat{k}. \]

Calculating the cross product:

\[ \vec{n} = \vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 0 & -4 \\ 3 & -1 & 5 \end{vmatrix}. \]

Calculating the determinant:

\[ \vec{n} = \hat{i}(0 \cdot 5 - (-4)(-1)) - \hat{j}(4 \cdot 5 - (-4)(3)) + \hat{k}(4 \cdot (-1) - 0 \cdot 3). \]

\[ = \hat{i}(0 - 4) - \hat{j}(20 + 12) + \hat{k}(-4 - 0), \]
\[ = -4 \hat{i} - 32 \hat{j} - 4 \hat{k}. \]

### Step 4: Write the equation of the plane
The equation of the plane can be expressed as:

\[ \vec{r} \cdot \vec{n} = \vec{P} \cdot \vec{n}. \]

Calculating \(\vec{P} \cdot \vec{n}\):

\[ \vec{P} \cdot \vec{n} = (1 \hat{i} + 1 \hat{j} + 1 \hat{k}) \cdot (-4 \hat{i} - 32 \hat{j} - 4 \hat{k}) = -4 - 32 - 4 = -40. \]

Thus, the equation of the plane is:

\[ \vec{r} \cdot (-4 \hat{i} - 32 \hat{j} - 4 \hat{k}) = -40. \]

### Step 5: Convert to Cartesian form
This can be written in Cartesian form as:

\[ -4x - 32y - 4z = -40. \]

Dividing through by -4 gives:

\[ x + 8y + z = 10. \]

### Step 6: Verify if the second line lies in the plane
The second line is given by:

\[ \vec{r} = (-\hat{i} + 2 \hat{j} + 5 \hat{k}) + \lambda (\hat{i} - 2 \hat{j} - 5 \hat{k}). \]

We can check if the point \((-1, 2, 5)\) lies on the plane:

Substituting into the plane equation:

\[ -1 + 8(2) + 5 = -1 + 16 + 5 = 20 \neq 10. \]

Thus, the second line does not lie in the plane.

### Final Answer
The equation of the plane is:

\[ x + 8y + z = 10. \]