If the line, {{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}} lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, {{x - 1} \over {12}} = {y \over 9} = {z \over 4} is :

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Answer: C. 0 Point (3, - 2, - \lambda ) on p line 2x - 4y + 3z - 2 = 0 = 6 + 8 - 3\lambda - 2 = 0 = 3\lambda = 12 \lambda = 4 Now, {{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1} . . .(i) {{x - 1} \over {12}} = {y \over 9} = {z \over 4} = {k_2} . . .(ii) Point on equation (i) P (k1 + 3, - k1 - 2, - 2k1 - 4) Point on equation (ii) Q(12k2 + 1, 9k2, 4k2) k1 + 3 = 12k2 + 1 \left| { - {k_1} - 2 = 9{k_2}} \right| - 2k1 - 4 = 4k2 k2 = 0 k1 = - 2 p (1, 0, 0) lie on equation of a line 1 gives shortest distance = 0

If the line, $${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}$$ lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, $${{x - 1} \over {12}} = {y \over 9} = {z \over 4}$$ is :

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