The lengths of the altitudes of a triangle $ ABC $ are:
$$ h_a = \frac{ 33 \sqrt{ 89}}{ 89 } ~~,~~ h_b = \frac{ 11 \sqrt{ 5}}{ 5 } ~~,~~ h_c = \frac{ 33 \sqrt{ 26}}{ 26 } $$The altitude $ h_a $ can be found using formula:
$$ A = \frac{ a \cdot h_a }{2} $$Where $ A $ is a triange area and $ a $ is a length of a side $ BC $. In this eample we have: $ A = \frac{ 33 }{ 2 } $ and $ d(B,C) = \sqrt{ 89 } $ so:
$$ A = \frac{ a \cdot h_a }{2} => h_a = \frac{2 \cdot A}{ a } = \frac{ 2 \cdot \frac{ 33 }{ 2 }}{ \sqrt{ 89 } } = \frac{ 33 \sqrt{ 89}}{ 89 } $$