Equation of the line passing through P(1,4,3) is: ‌
x−1
a
=‌
y−4
b
=‌
z−3
c
‌‌...... (i) Since equation (i) is perpendicular to ‌
x−1
2
=‌
y+3
1
=‌
z−2
4
and ‌
x+2
3
=‌
y−4
2
=‌
z+1
−2
Hence 2a+b+4c=0 and 3a+2b−2c=0∴‌‌‌
a
−2−8
=‌
b
12+4
=‌
c
4−3
⇒‌
a
−10
=‌
b
16
=‌
c
1
Hence the equation of the lines is ‌
x−1
−10
=‌
y−4
16
=‌
z−3
1
‌‌........ (ii) Now any point Q on (2) can be taken as (1−10λ,16λ+4,λ+3) ∴ Distance of Q from P(1,4,3)=(10λ)2+(16λ)2+λ2=357 ⇒‌‌(100+256+1)λ2=357 ⇒‌‌λ=1 or -1 ∴Q is (-9,20,4) or (11,-12,2) Hence a1+a2+a3=15 or 1