Math, asked by Preetbrar3039, 11 months ago

Let P be the plane, which contains the line of intersection of the planes, x + y + z - 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy – plane. Then the distance of the point (0, 0, 256) from P is equal to:
(A) 63√5 (B) 205√5
(C) 17/√5
(D) 11/√5

Answers

Answered by Anonymous
23

Answer:

Let P be the plane, which contains the line of intersection of the planes, x + y + z - 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy – plane. Then the distance of the point (0, 0, 256) from P is equal to:

(A) 63√5

(B) 205√5

(C) 17/√5

(D) 11/√5

Answered by Anonymous
7

Answer:

Let P be the plane, which contains the line of intersection of the planes, x + y + z - 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy – plane. Then the distance of the point (0, 0, 256) from P is equal to:

(A) 63√5

(B) 205√5✔

(C) 17/√5

(D) 11/√5

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