Let `vecx and vecy` are 2 non - zero and non - collinear vectors, then the largest value of k such that the non - zero vectors `(k^(2)-5k+6)vecx+(k-3)vecy` and `2vecx+5vecy` are collinear is

Options

A. 3
B. 6
C. `(12)/(5)`
D. `-1`

Answer

To solve the problem, we need to find the largest value of \( k \) such that the vectors \( (k^2 - 5k + 6)\vec{x} + (k - 3)\vec{y} \) and \( 2\vec{x} + 5\vec{y} \) are collinear.

### Step-by-step Solution:

1. **Understanding Collinearity**:
Two vectors \( \vec{a} \) and \( \vec{b} \) are collinear if there exists a scalar \( \lambda \) such that \( \vec{a} = \lambda \vec{b} \).

2. **Set Up the Equation**:
We can express the condition for collinearity as:
\[ (k^2 - 5k + 6)\vec{x} + (k - 3)\vec{y} = \lambda (2\vec{x} + 5\vec{y}) \]

3. **Equate Coefficients**:
From the equation above, we can equate the coefficients of \( \vec{x} \) and \( \vec{y} \):
- Coefficient of \( \vec{x} \):
\[ k^2 - 5k + 6 = 2\lambda \]
- Coefficient of \( \vec{y} \):
\[ k - 3 = 5\lambda \]

4. **Express \( \lambda \)**:
From the second equation, we can express \( \lambda \):
\[ \lambda = \frac{k - 3}{5} \]

5. **Substitute \( \lambda \)**:
Substitute \( \lambda \) into the first equation:
\[ k^2 - 5k + 6 = 2\left(\frac{k - 3}{5}\right) \]
Simplifying this gives:
\[ k^2 - 5k + 6 = \frac{2(k - 3)}{5} \]
Multiplying through by 5 to eliminate the fraction:
\[ 5(k^2 - 5k + 6) = 2(k - 3) \]
Expanding both sides:
\[ 5k^2 - 25k + 30 = 2k - 6 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 5k^2 - 25k - 2k + 30 + 6 = 0 \]
Simplifying further:
\[ 5k^2 - 27k + 36 = 0 \]

7. **Factoring the Quadratic**:
We can factor the quadratic equation:
\[ 5k^2 - 12k - 15k + 36 = 0 \]
Grouping terms:
\[ k(5k - 12) - 3(5k - 12) = 0 \]
Factoring out \( (5k - 12) \):
\[ (5k - 12)(k - 3) = 0 \]

8. **Finding the Roots**:
Setting each factor to zero gives:
\[ 5k - 12 = 0 \quad \Rightarrow \quad k = \frac{12}{5} \]
\[ k - 3 = 0 \quad \Rightarrow \quad k = 3 \]

9. **Determine the Largest Value**:
The possible values of \( k \) are \( \frac{12}{5} \) and \( 3 \). The largest value is:
\[ k = 3 \]

### Final Answer:
The largest value of \( k \) such that the vectors are collinear is \( \boxed{3} \).