Let α x+β y+y z=1 be the equation of a plane passing through the point (3,-2,5) and perpendicular to the line joining the points (1,2,3) and (-2,3,5). Then the value of α β y is equal to

Question

Let α x+β y+y z=1 be the equation of a plane passing through the point (3,-2,5) and perpendicular to the line joining the points (1,2,3) and (-2,3,5). Then the value of α β y is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given Equation is not equation of plane as

yy

is present. If we consider

y

is

γ

then answer would be 6 .
Normal vector of plane

= 3 \hat{i} - \hat{j} - 2 \hat{k}

Plane

: 3 x - y - 2 z + λ = 0

Point

(3, - 2, 5)

satisfies the plane

λ = - 1

3 x - y - 2 z = 1

α β y = 6

Q. Let αx+βy+yz=1 be the equation of a plane passing through the point (3,−2,5) and perpendicular to the line joining the points (1,2,3) and (−2,3,5). Then the value of αβy is equal to ____

Given Equation is not equation of plane as yy is present. If we consider y is γ then answer would be 6 . Normal vector of plane =3i^−j^​−2k^ Plane :3x−y−2z+λ=0 Point (3,−2,5) satisfies the plane λ=−1 3x−y−2z=1 αβy=6

Q. Let $αx + β y + yz = 1$ be the equation of a plane passing through the point $(3, - 2, 5)$ and perpendicular to the line joining the points $(1, 2, 3)$ and $(- 2, 3, 5)$ . Then the value of $α β y$ is equal to ____

Given Equation is not equation of plane as $yy$ is present. If we consider $y$ is $γ$ then answer would be 6 .
Normal vector of plane $= 3 \hat{i} - \hat{j} - 2 \hat{k}$
Plane $: 3 x - y - 2 z + λ = 0$
Point $(3, - 2, 5)$ satisfies the plane
$λ = - 1$
$3 x - y - 2 z = 1$
$α β y = 6$