Three Dimensional Geometry

Normal to plane is n= (-4i+5j+7k).
Plane: -4 (x-3)+5 (y-1)+7 (z-1)=0 ⇒ -4x+5y+7z=0.
Passes through (α, -3,5) ⇒ -4α-15+35=0 ⇒ α=5.

Two points on the line (say) x/3 = y/2, z=1 are (0,0,1) and (3,2,1)
So dr's of the line is (3,2,0)
Line passing through (1,2,1), parallel to L and coplanar with given plane is r = i+2j+k + t (3i+2j), t∈R (-2,0,1) satisfies the line (for t=-1)
⇒ (-2,0,1) lies on given plane.
Answer of the question is (B)
We can check other options by finding equation of plane
Equation plane: |x-1, y-2, z-1; 1+2, 2-0, 1-1; 2+2, 1-0, 2-1| = 0
⇒ 2 (x-1)-3 (y-2)-5 (z-1)=0
⇒ 2x-3y-5z+9=0

p∨ (¬p∧q) = (p∨¬p)∧ (p∨q)=T∧ (p∨q)=p∨q. Negation is ¬p∧¬q.

L: x/1=y/0=z/-1. Let a point on L be (r,0, -r).
Direction ratio of PN is (r-1, -2, -r+1).
PN is perpendicular to L, so (r-1) (1)+ (-2) (0)+ (-r+1) (-1)=0 ⇒ 2r-2=0 ⇒ r=1. N (1,0, -1).
Let Q be (s,0, -s). Direction ratio of PQ is (s-1, -2, -s+1).
PQ is parallel to plane x+y+2z=0, so normal to plane is perp to PQ.
(s-1) (1)+ (-2) (1)+ (-s+1) (2)=0 ⇒ s-1-2-2s+2=0 ⇒ -s-1=0 ⇒ s=-1. Q (-1,0,1).
PN = (0, -2,0). PQ= (-2, -2,2).
cosα = |PN.PQ|/|PN|PQ| = 4/ (2√12) = 1/√3.

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4

Line are coplanar
so | [α, 5-α, 1], [2, -1, 1], | = 0
−5α + (α – 5)3 + 7 = 0
-2α = 8 ⇒ α = −4
⇒ L? : (x+2)/-4 = (y+1)/9 = (z+1)/1
Now by cross checking option (A) is correct.

w = I + j - 3k
p = 3i - j + k
q = -3i + 2j + 4k
p * q = |i, j, k; 3, -1, 1; -3, 2, 4| = -6i - 15j + 3k
S.D. = |AD. (p * q)|/|p * q| = | (36+225+9)|/√ (36+225+9) = 3√30

Determinant of vectors must be zero. Vector between points on lines: (-1-k, -2-2, -3-3). Vector directions: (1,2,3) and (3,2,1).
| -1-k, -4, -6; 1, 2, 3; 3, 2, 1 | = 0.
(-1-k) (2-6) - (-4) (1-9) + (-6) (2-6) = 0.
4 (1+k) - 32 + 24 = 0.
4+4k - 8 = 0. 4k=4 ⇒ k=1.

Vector on plane: (3-2, 7-3, -7- (-2) = (1,4, -5).
Line direction vector (-3,2,1).
Normal to plane n = (1,4, -5)* (-3,2,1) = (14,14,14) or (1,1,1).
Plane: 1 (x-3)+1 (y-7)+1 (z+7)=0 ⇒ x+y+z-3=0.
d = |-3|/√3 = √3. d²=3.

Normal to the required plane is perpendicular to the normals of the given planes.
n = n?  * n?  = (2i + j - k) * (i - j - k) = -2i + j - 3k.
Equation of the plane is -2 (x+1) + 1 (y-0) - 3 (z+2) = 0
-2x - 2 + y - 3z - 6 = 0
-2x + y - 3z - 8 = 0
2x - y + 3z + 8 = 0
Comparing with ax + by + cz + 8 = 0, we get a=2, b=-1, c=3.
a+b+c = 2-1+3 = 4.