Which of the following is correct to the line $5 x + y + 3 z = 0$ and $3 x - y + z + 1 = 0$

Moderate

A

Symmetrical form of the equation of line is $\frac{x}{2} = \frac{y - \frac{1}{8}}{- 1} = \frac{z + \frac{5}{8}}{1}$
B

Symmetrical form of the equations of line is $\frac{x + \frac{1}{8}}{1} = \frac{y - \frac{5}{8}}{1} = \frac{z}{- 2}$
C

Equation of the plane through $(2, - 1, 4)$ and perpendicular to the given line is $2 x - y + z - 7 = 0$
D

Equation of line through $(2, - 1, 4)$ and perpendicular to the given line is $x + y - 2 z + 5 = 0$

Solution

The given planes are $5 x + y + 3 z = 0$ and $3 x - y + z + 1 = 0$
Suppose that the direction ratios of line of intersection of the above two planes is $⟨ a, b, c ⟩$
Hence, $3 a - b + c = 0, 5 a + b + 3 c = 0$ $\Rightarrow \frac{a}{1} = \frac{b}{1} = \frac{c}{- 2}$
To get a point on the line, substitute $z = 0$ in the plane equations and the solve for the other variables
hence a point on the line of intersection of planes is $P (- \frac{1}{8}, \frac{5}{8}, 0)$
Therefore, the equation of the required line is $\frac{x + \frac{1}{8}}{1} = \frac{y - \frac{5}{8}}{1} = \frac{z}{- 2}$

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