Consider two planes $P_{1} : 2 x - 3 y + z = - 5$ and . $P_{2} : x - 4 y - 14 z = 5$ . If a line L which is perpendicular to line of intersection of planes $P_{1} a n d P_{2}$ whose one point $A (- 7, y_{1}, z_{1})$ lies on the plane $P_{1}$ and point $B (x_{2}, y_{2}, z_{2})$ and $C (x_{3}, y_{3}, z_{3})$ lie on plane $P_{2}$ has the equation $\frac{x + 7}{a} = \frac{y - y_{1}}{- b} = \frac{z - z_{1}}{c}$ , where $a, b, c \in N$ , then:

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Least value of $(a + b + c)$ is 5

$y_{1} + z_{1} i s e q u a l t o - 4$

$\frac{z_{3} - z_{2}}{y_{3} - y_{2}} = 3$

$\frac{x_{3} - x_{2}}{z_{3} - z_{2}} = 2$

answer is D.

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Detailed Solution

Planes $P_{1} a n d P_{2}$ are perpendicular $L : \frac{x + 7}{a} = \frac{y - y_{1}}{- b} = \frac{z - z_{1}}{c}$

The above line lies in plane $P_{2}$ and perpendicular to plane $P_{1}$ $∴ A (- 7, y_{1}, z_{1})$
lies on both the planes
$∴ - 14 - 3 y_{1} + z_{1} = - 5 ...... (i)$
$- 7 - 4 y_{1} - 14 z_{1} = 5 .......... (i i)$
Solving equation (i) and (ii)
We get $y_{1} = - 3 a n d z_{1} = 0$ $∴ L : \frac{x + 7}{2} = \frac{y + 3}{- 3} = \frac{z - 0}{1}$

Now, verify the options.