The equation of the plane which passes throug | vedclass

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(A) The direction ratios of the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ are given by $(2-3, -1-4, 5-(-1)) = (-1, -5, 6)$.Since the plane

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$x + 5y - 6z + 19 = 0$

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(A) The direction ratios of the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ are given by $(2-3, -1-4, 5-(-1)) = (-1, -5, 6)$.Since the plane is normal to this line,the normal vector to the plane is $\vec{n} = (-1, -5, 6)$.The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the point $(2, -3, 1)$ and the normal vector $(-1, -5, 6)$,we get:$-1(x-2) - 5(y+3) + 6(z-1) = 0$$-x + 2 - 5y - 15 + 6z - 6 = 0$$-x - 5y + 6z - 19 = 0$Multiplying by $-1$,we get $x + 5y - 6z + 19 = 0$.

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by:

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