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The equations of given lines are
                                                          ...(1)

and                                 ...(2)

Any line through P (– 1, 3, –2) is
                                        ...(3)
where l, m, n are direction ratios of the line     
Since (3) is perpendicular to (1) and (2)
∴ l + 2 m + 3 n = 0    ...(4)
and – 3 l + 2 m + 5 n = 0    ...(5)
Solving (4) and (5), we get,
          

∴ from (3), the equations of line are

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The equations of the lines are
                        x = ay + b,    z = cy + d
and                  
or                  
and            
∴    direction ratios of two lines are a, 1, c and a', 1, c'.
The two lines are perpendicular if
(a) (a') + (1) (1) + (c) (c') = 0    [∵ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0]
i.e. if aa' + cc' + 1 = 0

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The equations of two lines are
                           

or                        
Direction ratios of two lines are

∵   lines are at right angle
         

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Given that a I + b m + c n = 0,
i.e.                                                ...(1)
Also,                                 ...(2)
Substituting value of n from (1) in (2), we get
              
or          
On dividing both sides by m², we have
     
If l₁, m₁, n₁ and l₂, m₂ , n₂ are the direction cosines of the two lines, then the roots of the equation (3) as 

                                   ...(4)
Similarly, using (1) and (2) and by elimination of 1, we get
                                                           ...(5)
Combining (4) and (5), we have
                       

                 ...(6)
We know that the lines with direction cosines l₁, m₁, n₁ and l₂, m₂, n₂ are perpendicular if
l₁ l₂ + m₁m₂ + n₁n₂ = 0    ....(7)
From (6) and (7), we get,

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The equations of given lines are
                                                            ...(1)
and                                                      ...(2)
Any line through  (α, β, γ) is

where l, m, n are direction-ratios of the line
Since (3) is perpendicular to (1)
∴ l l₁ + m m₁ + n n₁ = 0    ....(4)
Since (3) is perpendicular to (2)
∴ I l₂ + m m₂ + n n₂ = 0
Solving (4) and (5), we get,

∴   from (3), the equations of line are