Calling
vec v_1, vec v_2, vec v_3 the unit vectors with the direction of A-P, B-P and C-P we have
A = P + a vec v_1
B = P + b vec v_2
C = P + c vec v_3
Here
a = 10
b = 6
then
B-A = b vec v_2-a vec v_1
C-B = c vec v_3 - b vec v_2
but B-A and C-B are perpendicular then
<< B-A, C-B >> = << b vec v_2-a vec v_1, c vec v_3 - b vec v_2 >> = 0
b c << vec v_2, vec v_3 >> - b^2 norm(vec v_2)^2-ac << vec v_1, vec v_3 >>+a b << vec v_1, vec v_2 >> =0
and then
c = (b^2-ab << vec v_1, vec v_2>>)/(b << vec v_2, vec v_3 >> - a << vec v_1, vec v_3>>) = 33
with
<< vec v_1, vec v_2 >> =-1/2
<< vec v_2, vec v_3 >> =-1/2
<< vec v_1, vec v_3 >> = -1/2
NOTE
<< cdot, cdot >> represents the scalar product of two vectors.
norm(cdot) represents the norm operation.