Maths

Let the equation of the plane passing through (1, 2, 3) be:
a (x - 1) + b (y - 2) + c (z - 3) = 0
The plane contains the y-axis, which has direction ratios (0, 1, 0).
Therefore, the normal to the plane must be perpendicular to the y-axis.
a (0) + b (1) + c (0) = 0 ⇒ b = 0
The equation becomes: a (x - 1) + c (z - 3) = 0
ax + cz = a + 3c
The plane also passes through the origin (0,0,0) since it contains the y-axis.
a (0) + c (0) = a + 3c ⇒ a + 3c = 0 ⇒ a = -3c
Substitute a = -3c into the plane equation:
-3c (x - 1) + c (z - 3) = 0
-3 (x - 1) + (z - 3) = 0
-3x + 3 + z - 3 = 0
-3x + z = 0 ⇒ 3x - z = 0

Kindly consider the following figure

Kindly consider then following figure

According to the question,  
  (7 * 4) + (n * 6) = 52
6n = 24 ⇒ n = 4

The system of equations has no solution if the determinant of the coefficient matrix is zero.
Δ = |k 1|
|1 k 1|
|1 k|
Δ = k (k² - 1) - 1 (k - 1) + 1 (1 - k) = 0
Δ = k³ - k - k + 1 + 1 - k = 0
⇒ k³ - 3k + 2 = 0 ⇒ (k - 1)² (k + 2) = 0
∴ k = -2, 1
If k = 1 then all the equations are identical (infinite solutions). Hence k = -2 for no solution.

dy/dx = xy - 1 + x - y = x (y + 1) - (y + 1) = (x - 1) (y + 1)
⇒ ∫ dy / (1 + y) = ∫ (x - 1) dx ⇒ ln (1 + y) = x²/2 - x + C
For y (0) = 0, c = 0. ∴ ln (1 + y) = x²/2 - x
1 + y = e^ (x²/2 - x)
y (1) = e^ (1/2 - 1) - 1 = e^ (-1/2) - 1

n (s) = 6² = 36
E = { (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), (2, 1), (2, 2), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 5), (5, 1), (5, 2), (5, 3), (7, 1)}
∴ n (E) = 17
Required prob. = n (E) / n (S) = 17 / 36

S-2, L-2, A, B, Y, U
Required = ²C? ⋅? C? ⋅ 4!/2! = 2 ⋅ 10 ⋅ 24/2 = 240

P (at least 2 show 3 or 5) =? C? (2/6)² (4/6)² +? C? (2/6)³ (4/6)¹ +? C? (2/6)?
= (384+128+16)/6? = 11/27
n=27
∴ expectation of number of times = np
= 27 ⋅ (11/27) = 11

2x-y+3=0
4x-2y+α=0 ⇒ 2x-y+α/2=0
6x-3y+β=0 ⇒ 2x-y+β/3=0
d? = |α/2-3|/√ (2²+1²) = 1/√5 ⇒ |α-6|=2 ⇒ α-6=2, -2 ⇒ α=8,4
d? = |β/3-3|/√ (2²+1²) = 2/√5 ⇒ |β-9|=6 ⇒ β-9=6, -6 ⇒ β=15,3
Sum of all value of α and β = 30.