Find the all the values of lamda such that `(x,y,z)!=(0,0,0)`and `x(hati+hatj+3hatk)+y(3hati-3hatj+hatk)+z(-4hati+5hatj)=lamda(xhati+yhatj+zhatk)`

Options

A. A. `-2,0`
B. B `0,-2`
C. C. `-1,0`
D. D. `0,-1`

Explanation

To solve the given problem, we need to find all values of \( \lambda \) such that \( (x, y, z) \neq (0, 0, 0) \) and the equation

\[ x(\hat{i} + \hat{j} + 3\hat{k}) + y(3\hat{i} - 3\hat{j} + \hat{k}) + z(-4\hat{i} + 5\hat{j}) = \lambda (x\hat{i} + y\hat{j} + z\hat{k}) \]

holds true.

### Step 1: Expand the left-hand side

We start by expanding the left-hand side of the equation:

\[ x(\hat{i} + \hat{j} + 3\hat{k}) = x\hat{i} + x\hat{j} + 3x\hat{k} \]

\[ y(3\hat{i} - 3\hat{j} + \hat{k}) = 3y\hat{i} - 3y\hat{j} + y\hat{k} \]

\[ z(-4\hat{i} + 5\hat{j}) = -4z\hat{i} + 5z\hat{j} \]

Combining these, we have:

\[ (x + 3y - 4z)\hat{i} + (x - 3y + 5z)\hat{j} + (3x + y)\hat{k} \]

### Step 2: Set coefficients equal

Now, we equate the coefficients of \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) on both sides of the equation:

1. Coefficient of \( \hat{i} \):
\[ x + 3y - 4z = \lambda x \tag{1} \]

2. Coefficient of \( \hat{j} \):
\[ x - 3y + 5z = \lambda y \tag{2} \]

3. Coefficient of \( \hat{k} \):
\[ 3x + y = \lambda z \tag{3} \]

### Step 3: Rearranging the equations

Rearranging each equation gives us:

1. \( (1 - \lambda)x + 3y - 4z = 0 \)
2. \( x - (3 + \lambda)y + 5z = 0 \)
3. \( 3x + y - \lambda z = 0 \)

### Step 4: Forming the determinant

For the system of equations to have a non-trivial solution (i.e., \( (x, y, z) \neq (0, 0, 0) \)), the determinant of the coefficients must be zero:

\[ \begin{vmatrix} 1 - \lambda & 3 & -4 \\ 1 & - (3 + \lambda) & 5 \\ 3 & 1 & -\lambda \end{vmatrix} = 0 \]

### Step 5: Calculate the determinant

Calculating the determinant:

\[ = (1 - \lambda) \begin{vmatrix} -(3 + \lambda) & 5 \\ 1 & -\lambda \end{vmatrix} - 3 \begin{vmatrix} 1 & 5 \\ 3 & -\lambda \end{vmatrix} - 4 \begin{vmatrix} 1 & -(3 + \lambda) \\ 3 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( = (1 - \lambda)((-3 - \lambda) \cdot (-\lambda) - 5) \)
2. \( = -3(1 \cdot (-\lambda) - 15) \)
3. \( = -4(1 - 3(3 + \lambda)) \)

### Step 6: Set the determinant to zero

After simplifying and combining all terms, we set the determinant equal to zero and solve for \( \lambda \):

\[ \lambda^3 + 2\lambda^2 + \lambda = 0 \]

Factoring out \( \lambda \):

\[ \lambda(\lambda^2 + 2\lambda + 1) = 0 \]

This gives us:

\[ \lambda(\lambda + 1)^2 = 0 \]

### Step 7: Find the values of \( \lambda \)

From this equation, we find:

\[ \lambda = 0 \quad \text{or} \quad \lambda = -1 \]

### Final Answer

Thus, the values of \( \lambda \) such that \( (x, y, z) \neq (0, 0, 0) \) are:

\[ \lambda = 0 \quad \text{and} \quad \lambda = -1 \]