Let the foot of the perpendicular in the 2x−3y+4z=29 be P(α,β,γ) So, the point (α,β,γ) satisfy the given plane. ∴‌‌2α−3β+4γ=29....(i) Now DR's of PO is (α,β,γ). where O is origin. since, OF is perpendicular to the given plane Therefore, normal to the plane is parallel to OF.
∴‌‌‌
α
2
=‌
β
−3
=‌
γ
4
=k ⇒‌‌α=2k,β=−3k and γ=4k On putting the value of α,β and γ in Eq. (i), we get 2(2k)−3(−3k)+4(4k)=29 ⇒‌‌4k+9k+16k=29 ⇒‌‌29k=29⇒k=1 ∴α=2,β=−3 and γ=4 Hence, foot of perpendicular is (2,−3,4) .