Reqd. pt. #P in L={(1,0,-2)+lambda(1,2,2) | lambda in RR}#

#rArr EE t in RR, s.t., P=P(1+t,2t,2t-2)#

Now, #A(1,3,4), P(1+t,2t,2t-2) rArr vec(AP)=(t,2t-3,2t-6)#.

Since, #vec(AP) bot L rArr vec(AP) bot "the direction vector of" L#

#rArr vec(AP) bot (1,2,2) rArr vec(AP).(1,2,2)=0#

#:. (t,2t-3,2t-6).(1,2,2)=0. :. t+4t-6+4t-12=0#

#:. t=2 rArr P(1+t,2t,2t-2)=P(3,4,2)#.

As regards #Q in L#, like #P#, #Q# has to be #Q(1+s,2s,2s-2)# for

some #s in RR#

Because in right-isosceles #DeltaAPQ, m/_APQ=90^@, AP=PQ#.

With #A(1,3,4), P(3,4,2), Q(1+s,2s,2s-2), and, AP^2=PQ^2#, we have,

#(1-3)^2+(3-4)^2+(4-2)^2=(1+s-3)^2+(2s-4)^2+(2s-2-2)^2#

#:.4+1+4=(s-2)^2+4(s-2)^2+4(s-2)^2#.

#:. 9(s-2)^2=9 rArr s-2=+-1 rArr s=3, or, s=1#.

Accordingly, we have #2# pts. #Q_1, Q_2 in L# satisfying the given cond., namely,

for #s=3, Q_1(1+s,2s,2s-2)=Q_1(4,6,4), &, "for" s=1, Q_2(2,2,0)#.

Thus, the Reqd. Pts. are #P(3,4,2), Q_1(4,6,4), and, Q_2(2,2,0)#, as

obtained by Respected **Cesareo R.**

Enjoy Maths.!.