If the equation of a plane P, passing through the intersection of the planes, x+4 y-z+7=0 and 3 x+y+5 z=8 is a x+b y+6 z=15 for some a, b ∈ R, then the distance of the point (3,2,-1) from the plane P is .

Question

If the equation of a plane P, passing through the intersection of the planes, x+4 y-z+7=0 and 3 x+y+5 z=8 is a x+b y+6 z=15 for some a, b ∈ R, then the distance of the point (3,2,-1) from the plane P is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

D1 = |-7&4&-1 8&1&5 15&b&6|= 0 ⇒ b = -3 D = |1&4&-1 3&1&5 a&b&6| = 0 ⇒ 21 a - 8b - 66 = 0 dots (1) P: 2x - 3y + 6z = 15 so required distance = (21/7) = 3

Q. If the equation of a plane P, passing through the intersection of the planes, $x + 4 y - z + 7 = 0$ and $3 x + y + 5 z = 8$ is $a x + b y + 6 z = 15$ for some $a, b \in R,$ then the distance of the point (3,2,-1) from the plane $P$ is _______.

$D_{1} = ∣ ∣ - 7 815 41 b - 1 56 ∣ ∣ = 0$
$\Rightarrow b = - 3$
$D = ∣ ∣ 13 a 41 b - 1 56 ∣ ∣ = 0$
$\Rightarrow 21 a - 8 b - 66 = 0 \dots (1)$
$P : 2 x - 3 y + 6 z = 15$
so required distance $= \frac{21}{7} = 3$

Q. If the equation of a plane P, passing through the intersection of the planes, x+4y−z+7=0 and 3x+y+5z=8 is ax+by+6z=15 for some a,b∈R, then the distance of the point (3,2,-1) from the plane P is _______.

D1​=∣∣​−7815​41b​−156​∣∣​=0 ⇒b=−3 D=∣∣​13a​41b​−156​∣∣​=0 ⇒21a−8b−66=0…(1) P:2x−3y+6z=15 so required distance =721​=3

Q. If the equation of a plane P, passing through the intersection of the planes, $x + 4 y - z + 7 = 0$ and $3 x + y + 5 z = 8$ is $a x + b y + 6 z = 15$ for some $a, b \in R,$ then the distance of the point (3,2,-1) from the plane $P$ is _______.

$D_{1} = ∣ ∣ - 7 815 41 b - 1 56 ∣ ∣ = 0$
$\Rightarrow b = - 3$
$D = ∣ ∣ 13 a 41 b - 1 56 ∣ ∣ = 0$
$\Rightarrow 21 a - 8 b - 66 = 0 \dots (1)$
$P : 2 x - 3 y + 6 z = 15$
so required distance $= \frac{21}{7} = 3$