Three Dimensional Geometry

For point of intersection
1 + 3λ = 3 + μ
2 + λ = 1 + 2μ
5λ = 5 ⇒ λ = 1, μ = 1
Point of intersection (4, 3, 5)
For the greatest distance from origin perpendicular from meet plane at point of intersection
Hence equation r . (4i + 3j + 5k) = 50

$\lambda =1+\mu$

$\begin{array}{l}\underset{\_}{2\lambda =3\lambda \mu }\\ \lambda =2\to S\left(2,4,6\right)\end{array}$

$\mu =1$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \end{array}\right|=\left(7,-2,-1\right)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

$limPQ:\frac{x-1}{1}=\frac{y-0}{1}=\frac{3-1}{1}=\lambda$  

              $M\left(1+\lambda ,\lambda ,1+\lambda \right)$

x + y + z = 5

$\Rightarrow \lambda =1$  

M (2, 1, 2)

Q (3, 2, 3)

$L:\frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}=t$  

 R (3, 1, 1), QR² = 1 + 4 = 5

Let direction ratio of the normal to the required plane are l, m, n

$\begin{array}{l}\therefore 3l+m-2n=0\\ \therefore 2l-5m-n=0\\ \therefore \frac{l}{-11}=\frac{m}{-1}=\frac{n}{-17}\end{array}$             

Equation of required plane

11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0

$\Rightarrow 11x+y+17z+38=0$           

Any point on line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=r$ is P (r + 3, 2r + 4, 2r + 5) lies on x + y + z = 17,

5r + 12 = 17

r = 1

$P\left(4,6,7\right)A\left(1,1,9\right)distAP=\sqrt[]{9+25+4}=\sqrt[]{38}$       

| 1 -1 -1 |
| 1 -k | = 0
| k 2 1 |

⇒ 1 (1 + 2k) + 1 (1 + k²) – 1 (2 – k) = 0
2k + 1 + 1 + k² − 2 + k = 0
k² + 3k = 0
k = 0, -3

(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = -2 (3a-4b+12c+19)/ (3²+ (-4)²+12²)
(x-a)/3 = (y-b)/ (-4) = (z-c)/12 = (-6a+8b-24c-38)/169
(x, y, z) = (a–6, β, γ)
(a-b)-a)/3 = (β-b)/ (-4) = (γ-c)/12 = (-6a+8b-24c-38)/169
(β-b)/ (-4) = -2
=> β = 8+b
=> 3a – 4b + 12c = 150 . (i)
a + b + c = 5
=> 3a + 3b + 3c = 15 . (ii)
Applying (i) – (ii), we get :
= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

l + m – n = 0 => n = l + m
3l² + m² + cnl = 0
3l² + m² + cl (l+m) = 0
= (3+c) (l/m)² + c (l/m) + 1 = 0
? Lines are parallel
D = 0
c = 4 (as c > 0)

P will be the centroid of triangle ABC.
The centroid P is (x? +x? +x? )/3, (y? +y? +y? )/3).
The coordinates of P are given as (17/6, 8/3).
The coordinates of Q are not given, but a calculation is shown.
PQ = √ (24/6)² + (9/3)²) = √ (4² + 3²) = √ (16+9) = √25 = 5.
This implies the coordinates of Q are such that the difference in coordinates with P leads to this result. For example if P= (x? , y? ) and Q= (x? , y? ), then x? -x? =4 and y? -y? =3.