Test: Distance Of A Point From A Plane - JEE MCQ

 Length of perpendicular from (2,3,-5) to the plane x + 2y − 2z − 9 = 0.
= |(2 + 2×3 −2×(−5) − 9)|√1² + 2² + (−2)²
= |2 + 6 + 10 − 9|/√9
= 9/3
= 3 units.

Since, the foot of the perpendicular to the plane is A(2,4,6). Therefore (4,2,6) is the point on the plane.
So, equation of the plane passing through the point (2,4,6) is:
a(x–2)+b(y–4)+c(z–6)=0.
Now, the direction ratios of the perpendicular line OA=2+1, 4+3, 6+5, i.e., 3,7,11
Therefore, the required plane is: 
3(x–2)+7(y–4)+6(z–11)=0
i.e, 3x + 7y + 11z = 100

 Since the point is 3 units away from the horizontal plane the distance from the point to xy reference line will be 3 units. And then the point is at distance of 5 units from the vertical plane the distance from reference line and point will be 5, sum is 8.

(x + 1)/(-2) = (y - 2)/(3) = (z + 5)/(-6)
b = -2i + 3j - 6k
n =  2i – yj + 2k
Sinθ = (b.n)/|b||n|
Sinθ = (-2i + 3j - 6k).(2i – yj + 2k)]/{[(-2)² + (3)² + (-6)²]^½ *[(2)² + (-1)² + (2)²]^½}
Sinθ = (-4 -3 -12)/[(49)^½ * (9)^½]
Sinθ = 19/(7 * 3)
Sinθ= 19/21
θ = sin^-1(19/21)