Question 993287
Distance between P(a,3) and Q(6,2a) is 
{{{sqrt((6-a)^2 + (2a-3)^2)}}}
= {{{sqrt(a^2-12a+36 + 4a^2-12a+9)}}}
= {{{sqrt(5a^2-24a+45)}}}
For this distance to be greater than 29;
{{{sqrt(5a^2-24a+45) > 29}}}
=> {{{5a^2-24a+45 > 841}}}
=> {{{5a^2-24a-796 > 0}}}
Solving the quadratic we get roots as -10.44 and 15.24
Since coefficient of a^2 is positive, the graph is opening upwards
So the value of this quadratic is negative between these two roots.

So the required value of a for the quadratic to be positive, is 
All values of a except [-10.44,15.24] range.