Let the unit vector in the direction of the line be #l hat(i)+ m hat(j) +n hat(k)#. (The numbers #(l,m,n)# are also called the direction cosines of the line). Then, from the given data
#l = cos 120^circ = -1/2#
and
#m = cos 45 = 1/sqrt2#
Since #l^2+m^2+n^2 =1#, this gives
#m^2 = 1-l^2-n^2 = 1-(-1/2)^2-(1/sqrt2)^2 = 1/4#
Since #beta# is in the first quadrant, #m=cos beta >0# and so #m=1/2# and #beta = 60^circ#
Now, let #(x,y,z)# be the coordinates of any point on our line. Then the vector #(x-2) hati +(y+1) hatj + (z-5) hat k# joining the two points #(x,y,z)# to #(2,-1,5)# must be parallel to #l hat i +m hat j +n hatk = -1/2 hati + 1/2 hat j + 1/sqrt2 hat k#. Thus, there must be a number #t# such that
#(x-2) hati +(y+1) hatj + (z-5) hat k = t( -1/2 hati + 1/2 hat j + 1/sqrt2 hat k)#
Thus we arrive at the parametric equations
#x = 2-t/2,qquad y = -1+t/2,qquad y = 5+t/sqrt2#
Note
You could, of course parametrize the line in several different ways. For example, using #tau=t/2# as the parameter will change the equations to
#x = 2-tau,qquad y = -1+tau,qquad y = 5+sqrt2 tau#
