Show that the following pairs of lines intersect. Also find their point of intersection :
(i) `(x-1)/(2) = (y-2)/(3)=(z-3)/(4)` and `(x-4)/(5)=(y-1)/(2)=z`
(ii) `(x-4)/(1)=(y+3)/(-4) = (z+1)/(7)` and `(x-1)/(2) = (y+1)/(-3)=(z+10)/(8)`

Answer

To show that the given pairs of lines intersect and to find their point of intersection, we will solve each part step by step.

### Part (i)

Given lines:
1. \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\)
2. \(\frac{x-4}{5} = \frac{y-1}{2} = z\)

Let us denote the common ratio for the first line as \(r_1\) and for the second line as \(r_2\).

From the first line:
- \(x = 2r_1 + 1\)
- \(y = 3r_1 + 2\)
- \(z = 4r_1 + 3\)

From the second line:
- \(x = 5r_2 + 4\)
- \(y = 2r_2 + 1\)
- \(z = r_2\)

Now, we will set the expressions for \(x\), \(y\), and \(z\) from both lines equal to each other:

1. Setting \(x\) equal:
\[ 2r_1 + 1 = 5r_2 + 4 \]
Rearranging gives:
\[ 2r_1 - 5r_2 = 3 \quad \text{(Equation 1)} \]

2. Setting \(y\) equal:
\[ 3r_1 + 2 = 2r_2 + 1 \]
Rearranging gives:
\[ 3r_1 - 2r_2 = -1 \quad \text{(Equation 2)} \]

Now we will solve these two equations simultaneously.

From Equation 1:
\[ 2r_1 = 5r_2 + 3 \implies r_1 = \frac{5r_2 + 3}{2} \]

Substituting \(r_1\) into Equation 2:
\[ 3\left(\frac{5r_2 + 3}{2}\right) - 2r_2 = -1 \]
Multiplying through by 2 to eliminate the fraction:
\[ 3(5r_2 + 3) - 4r_2 = -2 \]
Expanding:
\[ 15r_2 + 9 - 4r_2 = -2 \]
Combining like terms:
\[ 11r_2 + 9 = -2 \implies 11r_2 = -11 \implies r_2 = -1 \]

Now substituting \(r_2 = -1\) back to find \(r_1\):
\[ r_1 = \frac{5(-1) + 3}{2} = \frac{-5 + 3}{2} = \frac{-2}{2} = -1 \]

Now substituting \(r_1 = -1\) into the equations for \(x\), \(y\), and \(z\):
\[ x = 2(-1) + 1 = -2 + 1 = -1 \]
\[ y = 3(-1) + 2 = -3 + 2 = -1 \]
\[ z = 4(-1) + 3 = -4 + 3 = -1 \]

Thus, the point of intersection is \((-1, -1, -1)\).

### Part (ii)

Given lines:
1. \(\frac{x-4}{1} = \frac{y+3}{-4} = \frac{z+1}{7}\)
2. \(\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8}\)

Let \(r_1\) and \(r_2\) be the common ratios for the first and second lines, respectively.

From the first line:
- \(x = r_1 + 4\)
- \(y = -4r_1 - 3\)
- \(z = 7r_1 - 1\)

From the second line:
- \(x = 2r_2 + 1\)
- \(y = -3r_2 - 1\)
- \(z = 8r_2 - 10\)

Setting the expressions for \(x\), \(y\), and \(z\) equal to each other:

1. Setting \(x\) equal:
\[ r_1 + 4 = 2r_2 + 1 \]
Rearranging gives:
\[ r_1 - 2r_2 = -3 \quad \text{(Equation 3)} \]

2. Setting \(y\) equal:
\[ -4r_1 - 3 = -3r_2 - 1 \]
Rearranging gives:
\[ -4r_1 + 3r_2 = 2 \quad \text{(Equation 4)} \]

Now we will solve these two equations simultaneously.

From Equation 3:
\[ r_1 = 2r_2 - 3 \]

Substituting \(r_1\) into Equation 4:
\[ -4(2r_2 - 3) + 3r_2 = 2 \]
Expanding:
\[ -8r_2 + 12 + 3r_2 = 2 \]
Combining like terms:
\[ -5r_2 + 12 = 2 \implies -5r_2 = -10 \implies r_2 = 2 \]

Now substituting \(r_2 = 2\) back to find \(r_1\):
\[ r_1 = 2(2) - 3 = 4 - 3 = 1 \]

Now substituting \(r_1 = 1\) into the equations for \(x\), \(y\), and \(z\):
\[ x = 1 + 4 = 5 \]
\[ y = -4(1) - 3 = -4 - 3 = -7 \]
\[ z = 7(1) - 1 = 7 - 1 = 6 \]

Thus, the point of intersection is \((5, -7, 6)\).

### Summary of Results
1. The point of intersection for the first pair of lines is \((-1, -1, -1)\).
2. The point of intersection for the second pair of lines is \((5, -7, 6)\).