The distance between two complex numbers z and v is given by |z - v|, where |.| denotes the modulus or absolute value.
Therefore, we have:
|z - v| = √26
Substituting z = 5 + ai and v = 2a - 3i, we get:
|5 + ai - (2a - 3i)| = √26
Simplifying the expression inside the modulus, we get:
|5 - 2a + (a + 3i)| = √26
|3 - a + 3i| = √26
Taking the modulus of the complex number on the left side, we get:
√((3-a)² + 3²) = √26
Squaring both sides, we get:
(3-a)² + 9 = 26
Expanding the left side and simplifying, we get:
a² - 6a - 14 = 0
We can solve this quadratic equation using the quadratic formula:
a = [6 ± √(6² + 4(14))]/2
a = [6 ± √52]/2
a = 3 ± √13
Therefore, the value(s) of a that satisfy the given condition are
a = 3 + √13 and a = 3 - √13.