Vector Algebra

Kindly consider the following figure

Given r * a = r * b, which means r * a - r * b = 0 ⇒ r * (a - b) = 0.
This implies that vector r is parallel to vector (a - b).
So, r = λ (a - b) for some scalar λ.
a - b = (2i - 3j + 4k) - (7i + j - 6k) = -5i - 4j + 10k.
So, r = λ (-5i - 4j + 10k).
We are also given r ⋅ (i + 2j + k) = -3.
λ (-5i - 4j + 10k) ⋅ (i + 2j + k) = -3
λ (-51 - 42 + 10*1) = -3
λ (-5 - 8 + 10) = -3
λ (-3) = -3 ⇒ λ = 1.
So, r = 1 * (-5i - 4j + 10k) = -5i - 4j + 10k.
We need to find r ⋅ (2i - 3j + k).
(-5i - 4j + 10k) ⋅ (2i - 3j + k) = (-5) (2) + (-4) (-3) + (10) (1)
= -10 + 12 + 10 = 12.

Volume of parallelepiped v = | [a? b? c? ]|
v = | 1 n |
| 2 4 -n| = ±158
| 1 n 3 |
1 (12+n²) - 1 (6+n) + n (2n-4) = ±158
3n²-5n+152=0 or 3n²-5n+164=0
D<0 (no real roots)
n=8, -19/3 ⇒ n=8
then b? ⋅c? = 2+4n-3n=10
a? ⋅c? = 1+n+3n=33

|x + y|² = |x|²
(x+y)· (x+y) = x·x
|x|² + 2x·y + |y|² = |x|²
|y|² + 2x·y = 0 (1)
and (2x + λy)·y = 0
2x·y + λ|y|² = 0 (2)
From (1), 2x·y = -|y|².
Substitute into (2):
-|y|² + λ|y|² = 0
(λ-1)|y|² = 0
Assuming y is a non-zero vector, |y|² ≠ 0, therefore λ=1.

Let a? = xî + y? + zk
î * (a? * î) = (î·î)a? – (a? ·î)î = y? + zk
Similarly? * (a? *? ) = xî + zk and k? * (a? * k? ) = xî + yk
|î * (a? * î)|² + |? * (a? *? )|² + |k? * (a? * k? )|²
= |y? + zk|² + |xî + zk|² + |xî + y? |² = 2|a? |² = 2 (9) = 18

f (x) = a? ⋅ (b? * c? ) = |x -2 3; -2 x -1; 7 -2 x|
= x³ - 27x + 26
f' (x) = 3x² - 27 = 0 ⇒ x = ±3 and f' (-3) < 0
⇒ local maxima at x = x? = -3
Thus, a? = -3i? - 2j? + 3k? , b? = 2i? - 3j? - k? , and c? = 7i? - 2j? - 3k?
⇒ a? ⋅ b? + b? ⋅ c? + c? ⋅ a? = 9 - 5 - 26 = -22