The distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0 measured parallel to the line $\frac{x + 3}{3} = \frac{y - 2}{6} = \frac{z}{2}$ is

Q. If

P_{1}

and

P_{2}

are the lengths of the perpendiculars from the points (2,3,4) and (1,1,4) respectively from the plane 3x-6y+2z+11 =0, then

P_{1}

and

P_{2}

are the roots of the equation

Join BYJU'S Learning Program

Distance of the point (2,3,4) from the plane 3x−6y+2z+11=0 is

The correct option is A 1unit Required distance is=∣∣ ∣ ∣∣3(2)−6(3)+2(4)+11√32+(−6)2+22∣∣ ∣ ∣∣=1

Distance of the point $(2, 3, 4)$ from the plane $3 x - 6 y + 2 z + 11 = 0$ is

The correct option is A $1 u n i t$
Required distance is
$= ∣ ∣ ∣ ∣ ∣ \frac{3 (2) - 6 (3) + 2 (4) + 11}{\sqrt{3^{2} + (- 6)^{2} + 2^{2}}} ∣ ∣ ∣ ∣ ∣ = 1$