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(D) Let the general point on the line be $P(6 - \lambda, -1, -3 + 4\lambda)$.Since this point lies on the plane $x + y - z = 3$,we substitute the coor

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$(5, -1, 1)$

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(D) Let the general point on the line be $P(6 - \lambda, -1, -3 + 4\lambda)$.Since this point lies on the plane $x + y - z = 3$,we substitute the coordinates into the plane equation:$(6 - \lambda) + (-1) - (-3 + 4\lambda) = 3$$6 - \lambda - 1 + 3 - 4\lambda = 3$$8 - 5\lambda = 3$$5\lambda = 5$$\lambda = 1$Substituting $\lambda = 1$ back into the point $P$,we get:$x = 6 - 1 = 5$$y = -1$$z = -3 + 4(1) = 1$Thus,the point of intersection is $(5, -1, 1)$.

Find the coordinates of the point of intersection of the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ and the plane $x + y - z = 3$.

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