x – y + 3z – 2 = 0 ; (3, 1, 0) ;

Sol. Let P(1, 2, 1) Q(1, 4, 2) and R(2, 3, 5). Now d.r.s. of the line QR are 2 – 1, 3 – 4, 5 – 2 = 1, – 1, 3

The reqd. plane is to pass through 'P' and ⊥ QR. Its equation is 1(x – 1) – 1 · (y – 2) + 3(z – 1) = 0

⇒ x – y + 3z – 2 = 0 ...(i)

Let B( α , β , γ ) be the foot of the ⊥ from A(4, 0, 3) on (1).

∴ α – β + 3 γ – 2 = 0 ...(ii)

and drs. of AB (Normal to the plane (i) are) α – 4, β – 0, γ – 3

Hence = = = k (say) ⇒ α = k + 4, β = – k, γ = 3k + 3 ...(iii)

From (ii) and (iii)

(k + 4) + k + 3(3k + 3) – 2 = 0 ⇒ k = –1

∴ The coordinates of the foot of ⊥ from A on (1) is B( α , β , γ ) = (3, 1, 0)

and ⊥ = AB = =

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