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(B) The direction ratios of the normal to the plane $x+y+3z-4=0$ are $(1, 1, 3)$.Since the perpendicular line passes through the origin $(0, 0, 0)$,it

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$\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)$

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(B) The direction ratios of the normal to the plane $x+y+3z-4=0$ are $(1, 1, 3)$.Since the perpendicular line passes through the origin $(0, 0, 0)$,its equation is $\frac{x}{1} = \frac{y}{1} = \frac{z}{3} = K$.Any point on this line is of the form $P(K, K, 3K)$.Since this point $P$ lies on the plane,it must satisfy the plane equation:$K + K + 3(3K) - 4 = 0$$2K + 9K = 4$$11K = 4 \Rightarrow K = \frac{4}{11}$.Substituting $K$ back into the coordinates of $P$,we get $P = \left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)$.

The foot of the perpendicular drawn from the origin to the plane $x+y+3z-4=0$ is

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