Let x-2/3 = y+1/-2 = z+3/-1 lie on the plane px – qy + z = 5, for s

1.	Let x-2/3 = y+1/-2 = z+3/-1 lie on the plane px – qy + z = 5, for some p, q ∈ R .The shortest distance of the plane from the origin is:(A) √(3/109)(B) √(5/142)(C) √(5/71)(D) √(1/142)
Answer» Correct option is (B) √(5/142) (2, –1, –3) satisfy the given plane. So 2p + q = 8 .... (i) Also given line is perpendicular to normal plane so 3p + 2q – 1 = 0 .... (ii) ⇒ p = 15, q = –22 Eq. of plane 15x – 22y + z – 5 = 0 its distance from origin = 6/√710 = √(5/142)

Let x-2/3 = y+1/-2 = z+3/-1 lie on the plane px – qy + z = 5, for some p, q ∈ R .The shortest distance of the plane from the origin is:(A) √(3/109)(B) √(5/142)(C) √(5/71)(D) √(1/142)

Answer»

Correct option is (B) √(5/142)

(2, –1, –3) satisfy the given plane.

So 2p + q = 8 .... (i)

Also given line is perpendicular to normal plane so

3p + 2q – 1 = 0 .... (ii)

⇒ p = 15, q = –22

Eq. of plane 15x – 22y + z – 5 = 0

its distance from origin

= 6/√710 = √(5/142)

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Let x-2/3 = y+1/-2 = z+3/-1 lie on the plane px – qy + z = 5, for some p, q ∈ R .The shortest distance of the plane from the origin is:(A) √(3/109)(B) √(5/142)(C) √(5/71)(D) √(1/142)

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