If the equation of a line is frac{x + 3}{2} = | vedclass

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(A) The direction ratios of the line are $(a_1, b_1, c_1) = (2, 3, 2)$.The normal vector to the plane $4x - 2y - z = 1$ has direction ratios $(a_2, b_

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Line is parallel to the plane

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(A) The direction ratios of the line are $(a_1, b_1, c_1) = (2, 3, 2)$.The normal vector to the plane $4x - 2y - z = 1$ has direction ratios $(a_2, b_2, c_2) = (4, -2, -1)$.For a line to be parallel to a plane,the dot product of the direction ratios of the line and the normal vector of the plane must be zero,i.e.,$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.Calculating the dot product: $(2)(4) + (3)(-2) + (2)(-1) = 8 - 6 - 2 = 0$.Since the dot product is $0$,the line is parallel to the plane.Next,we check if the line lies in the plane by testing a point on the line $(-3, 4, -5)$ in the plane equation: $4(-3) - 2(4) - (-5) = -12 - 8 + 5 = -15 
eq 1$.Since the point does not satisfy the plane equation,the line is strictly parallel to the plane.

If the equation of a line is $\frac{x + 3}{2} = \frac{y - 4}{3} = \frac{z + 5}{2}$ and the equation of a plane is $4x - 2y - z = 1$,then which of the following is true?

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