Key Concepts and Formulas

• A homogeneous second-degree equation of the form $Ax^2 + 2Hxy + By^2 = 0$ represents a pair of straight lines passing through the origin.
• The condition for these two lines to be perpendicular to each other is $A + B = 0$, where $A$ is the coefficient of $x^2$ and $B$ is the coefficient of $y^2$.

Step-by-Step Solution

Step 1: Identify the Coefficients of the Given Equation

We are given the equation $3a{x^2} + 5xy + \left( {{a^2} - 2} \right){y^2} = 0$. We need to identify the coefficients $A$, $2H$, and $B$ by comparing it to the general form $Ax^2 + 2Hxy + By^2 = 0$.

• The coefficient of $x^2$ is $A$. From the given equation, $A = 3a$.
• The coefficient of $xy$ is $2H$. From the given equation, $2H = 5$. Therefore, $H = \frac{5}{2}$. While $H$ is important for other conditions, it's not needed for the perpendicularity condition.
• The coefficient of $y^2$ is $B$. From the given equation, $B = a^2 - 2$.

Step 2: Apply the Perpendicularity Condition

For the lines to be perpendicular, the condition $A + B = 0$ must be satisfied. We substitute the expressions for $A$ and $B$ that we identified in Step 1.

$A + B = 0$ $(3a) + (a^2 - 2) = 0$

Step 3: Formulate and Solve the Quadratic Equation for 'a'

We rearrange the equation from Step 2 to form a standard quadratic equation in terms of $a$:

$a^2 + 3a - 2 = 0$

This is a quadratic equation of the form $pa^2 + qa + r = 0$, where $p=1$, $q=3$, and $r=-2$. We solve for $a$ using the quadratic formula:

$a = \frac{-q \pm \sqrt{q^2 - 4pr}}{2p}$

Substituting the values of $p$, $q$, and $r$, we get:

$a = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)}$ $a = \frac{-3 \pm \sqrt{9 + 8}}{2}$ $a = \frac{-3 \pm \sqrt{17}}{2}$

Step 4: Determine the Number of Values for 'a'

The quadratic formula yields two distinct real values for $a$:

$a_1 = \frac{-3 + \sqrt{17}}{2}$ $a_2 = \frac{-3 - \sqrt{17}}{2}$

Since we found two distinct real values of $a$ for which the lines are perpendicular, the correct option describes this scenario.

Common Mistakes & Tips

• Remember the Condition: The condition $A + B = 0$ is only valid for a homogeneous equation of degree 2 representing a pair of straight lines passing through the origin.
• Check Coefficients Carefully: Ensure you correctly identify the coefficients $A$, $2H$, and $B$ from the given equation, paying close attention to signs.
• Reality of Lines: The question only asks for perpendicularity, not whether the lines are real. The condition for real lines is $H^2 \ge AB$, but that's not required here.

Summary

The problem requires the direct application of the perpendicularity condition for a pair of lines represented by a homogeneous second-degree equation. By correctly identifying the coefficients $A$ and $B$ and applying the condition $A+B=0$, we arrive at a quadratic equation in $a$. Solving this quadratic equation yields two distinct values for $a$, meaning there are two values for which the given pair of lines will be perpendicular.

The final answer is $\boxed{\text{A}}$, which corresponds to option (A).