Find the value of k so that the lines  \(\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2k}} = \frac{{z - 3}}{2}\;and\frac{{7 - 7x}}{{3k}} = \frac{{5 - y}}{1} = \frac{{6 - z}}{5}\) are perpendicular to each other?

Concept:

If the angle between the lines \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\) where a1, b1, c1, a2, b2 and c2 are the direction ratios is 90° then

a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 0

Calculation:

Given: Equation of two lines is \(\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2k}} = \frac{{z - 3}}{2}\;and\frac{{7 - 7x}}{{3k}} = \frac{{5 - y}}{1} = \frac{{6 - z}}{5}\)

The given equation can be re-written as:

\(⇒ \frac{{x - 1}}{{ - \;3}} = \frac{{y - 2}}{{\frac{{2k}}{7}}} = \frac{{z - 3}}{2}\;\ and \ \frac{{x - 1}}{{\frac{{ - \;3k}}{7}}} = \frac{{y - 5}}{{ - \;1}} = \frac{{z - 6}}{{ - \;5}}\)

The direction ratios of the given lines are: 

⇒ a1 = - 3, b1 = 2k/7, c1 = 2, a2 = - 3k/7, b2 = - 1 and c2 = - 5

∵ The given lines are perpendicular to each to each other

As we know that if two lines are perpendicular to each to each other then a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 0

⇒ \({a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = \; - 3 \cdot \frac{{ - \;3k}}{7} + \left( {\frac{{2k}}{7}} \right) \cdot \left( { - \;1} \right) + 2 \cdot \left( { - \;5} \right) = 0\)

⇒ k = 10