Let A (1, – 1, – 3), B (4, – 3, 1), C (3, – 1, 2) be vertices of Δ ABC

Direction-ratios of AB are 4 – 1, – 3 + 1, 1 + 3 i.e.. 3, – 2, 4 respectively.
Direction-ratios of AC are 3 – 1, –1 + 1, 2 + 3 i.e., 2, 0, 5 respectively.

Area of 

Let A (1, 2, 4), B (– 2, 1, 2), C (2, 4, – 3) be vertices of Δ ABC.



Direction-ratios of AB are – 2, – 1, 1 – 2, 2 – 4 i.e., – 3, – 1,– 2 respectively.
Direction-ratios of AC are 2 – 1, 4 – 2, – 3 – 4 i.e., 1, 2, – 7 respectively.


                          

Tips: -

Note on parallelopiped and cube
(i) A parallelopipcd is a solid bounded by three pairs of parallel plane faces.
(ii) A rectangular parallelopiped is parallelopiped whose faces are all rectangles.
(iii) A cube is a parallelopiped whose faces are all squares.

The equations of given line are

From P (1, 6, 3) , draw PM ⊥ AB and produce it to P' (α, β γ) such that M is mid-point of PP'. Then P' is image of P in line AB.
Any point M as line AB is
(r. 2 r + 1, 3 r + 2)
Direction ratios of AB are 1, 2, 3
Direction ratios of PM are r – 1 , 2 r + 1 – 6 , 3 r + 2 – 3
i.e. r – 1, 2 r – 5, 3 r – 1
∵ PM ⊥ AB∴ (1) (r – 1) + (2) (2 r – 5) + (3) (3 r – 1) = 0
∴ i + 4 r – 10 + 9 r – 3 = 0
∴ 14 r = 14 ⇒ r = 1
∴ M is (1, 3, 5)
Now M is mid-point of PP'

Take O, a corner of cube OBLCMANP, as origin and OA, OB, OC, the three edges through it as the axes.

Let OA = OB = OC = a, then the co-ordinates of O , A , B , C are (0, 0, 0), (a, 0, 0), (0, a, 0), (0, 0, a) respectively ; those of P, L, M, N are (a, a, a), (0, a, a), (a, 0, a), (a, a, 0) respectively.
The four diagonals are OP, AL, BM, CN. Direction cosines of OP are proportional to a – 0, a – 0, a – 0, i.e., a, a, a, i.e., 1,1,1.
Direction-cosines of AL are proportional to 0 – a, a – 0, a – 0 i.e., –a, a, a, i.e., – 1, 1, 1.
Direction-cosines of BM are proportional to a – 0, 0 – a, a – 0, i.e., a – a, a i.e., 1, – 1, 1.
Direction-cosines of CN are proportional to a – 0, a – 0, 0 – a i.e., a, a,– a i.e., 1, 1, – 1.
  direction -cosines of OP are 
Directon-cosines of AL are 
Direction-cosines of BM are 
Direction-cosines of CN are 
Let l, m, n be direction-cosines of the line
∴    the line makes an angle α with OP.
              
or                                                          ...(1)
Similarly                                             ...(2)
                                                             ...(3)

Squaring and adding (1), (2), (3) and (4), we get,
cos² α + cos² β + cos² γ + cos² δ
                      

The equation of line BC
or   

Any point D on it is
(4 k + 1, 4, – 2 k + 6)
Let D be foot of perpendicular from A on BC.
Direction ratios of AD are
4 k + 1–1, 4 – 2, – 2 k + 6 – 1 i.e., 4 k , 2, – 2 k + 5
Direction ratios of BC are 4, 0, – 2
Since AD is perpendicular to BC
∴    (4 k) (4) + (2) (0) + (– 2 k + 5) (–2) = 0