Given, x+2y+z=6 . . . (i) and ‌‌y+2z=4 . . . (ii) Put y=4−2z from Eq. (ii) Eq. in (i), we get ‌x+8−4z+z‌‌=6 ⇒‌x‌‌=−2+3z ⇒‌‌
x+2
3
‌‌=z . . . (iii) ‌y‌‌=4−2z ⇒‌‌
y−4
−2
‌‌=z . . . (iv) From Eqs. (iii) and (iv), line of intersection of two planes is ‌
x+2
3
=‌
y−4
−2
=‌
z
1
=λ Then, Direction ratios of PQ is x−3λ−4−2λ,z=λ (3λ−5,−2λ+2,λ−1) (3λ−5,−2λ+2,λ−1)
Since, PQ is perpendicular to the line, then 3(3λ−5)−2(−2λ+2)+1(λ−1)=0 ∴‌‌λ=‌