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 :-

Question

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

Tardigrade Admin · Accepted Answer

λ(x+y+z-6) +2x+3y+z+5=0 (λ+2)x+(λ+3)y+(λ+1)z+5 -6λ=0 λ+1 =0 ⇒ λ=-1 P:x +2y +11 =0 =(11/√5)

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

$635$

$2055$

$17/ 5$

$11/ 5$

$λ (x + y + z - 6) + 2 x + 3 y + z + 5 = 0$
$(λ + 2) x + (λ + 3) y + (λ + 1) z + 5 - 6 λ = 0$
$λ + 1 = 0 \Rightarrow λ = - 1$
$P : x + 2 y + 11 = 0 = \frac{11}{5}$

Q. 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 :-

635​

2055​

17/5​

11/5​