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(D) Let the equation of the plane be $a(x - 4) + b(y - 3) + c(z - 2) = 0$. Since the plane contains the lines,the normal vector $(a, b, c)$ is perpend

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None of these

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(D) Let the equation of the plane be $a(x - 4) + b(y - 3) + c(z - 2) = 0$. Since the plane contains the lines,the normal vector $(a, b, c)$ is perpendicular to the direction vectors of both lines,$(1, 1, 2)$ and $(1, -4, 5)$. Thus,$a + b + 2c = 0$ and $a - 4b + 5c = 0$. Using the cross product to find the normal vector $(a, b, c) = (1, 1, 2) 	imes (1, -4, 5)$: $a = (1)(5) - (2)(-4) = 5 + 8 = 13$. $b = (2)(1) - (1)(5) = 2 - 5 = -3$. $c = (1)(-4) - (1)(1) = -4 - 1 = -5$. So the normal vector is $(13, -3, -5)$. The equation of the plane is $13(x - 4) - 3(y - 3) - 5(z - 2) = 0$. $13x - 52 - 3y + 9 - 5z + 10 = 0$. $13x - 3y - 5z - 33 = 0$ or $13x - 3y - 5z = 33$. Since this does not match any of the given options,the correct answer is $(d)$.

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{-4} = \frac{z}{5}$ is

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