Given:
The equation is x^2 - 2ax + a^2 + a - 3 = 0

To find:
The range in which a lies for the roots of the equation to be real and less than 3.

Explanation:

Let's solve the equation to find its roots:

x^2 - 2ax + a^2 + a - 3 = 0

The discriminant (D) of the quadratic equation is given by:

D = b^2 - 4ac

In this case, a = 1, b = -2a, and c = a^2 + a - 3. Substituting these values into the discriminant formula, we get:

D = (-2a)^2 - 4(1)(a^2 + a - 3)
= 4a^2 - 4(a^2 + a - 3)
= 4a^2 - 4a^2 - 4a + 12
= -4a + 12

For the roots to be real, the discriminant must be greater than or equal to zero:

D ≥ 0
-4a + 12 ≥ 0
-4a ≥ -12
a ≤ 3

So we have found that a must be less than or equal to 3 for the roots to be real.

Now, let's consider the condition that the roots should be less than 3.

Let the roots of the equation be α and β.

We know that in a quadratic equation, the sum of the roots (α + β) is given by:

α + β = -b/a

In this case, α + β = 2a/a = 2

The product of the roots (α * β) is given by:

α * β = c/a

In this case, α * β = (a^2 + a - 3)/a

Since the roots are real and less than 3, we have the following conditions:

α + β < />
2 < />
α * β < />
(a^2 + a - 3)/a < />

Simplifying the second inequality:

a^2 + a - 3 < />
a^2 - 2a - 3 < />

Now, let's find the range in which a lies by solving the quadratic inequality:

(a - 3)(a + 1) < />

The solutions are a < -1="" or="" a="" /> 3

However, we already know that a ≤ 3 from the discriminant condition. Therefore, the range in which a lies is:

a ≤ 3

This means that a must be less than or equal to 3 for the roots of the equation to be real and less than 3.