\(\mathrm{A}(0,-1,4), \mathrm{B}(1,2,3), \mathrm{C}(5,4,-1)\), then the foot of perpendicular from \(\mathrm{A}\) on \(\underline{\mathrm{BC}}\) is (A) \((-3,3,1)\) (B) \((3,-3,1)\) (C) \((3,3,1)\) (D) \((3,3,-1)\)

First, we need to calculate the direction ratios of the line segment BC using the coordinates of points B and C: Direction Ratios: \(m_x = C_x - B_x\), \(m_y = C_y - B_y\), and \(m_z = C_z - B_z\) Using the coordinates of points B \((1,2,3)\) and C \((5,4,-1)\), we can compute: \[m_x = 5 - 1 = 4\] \[m_y = 4 - 2 = 2\] \[m_z = -1 - 3 = -4\] So, the direction ratios of line segment BC are: \(m_x=4\), \(m_y=2\), and \(m_z=-4\).

Using the coordinates of point B and the direction ratios, construct the vector equation of the line segment BC: \[ \frac{x - x_B}{m_x} = \frac{y - y_B}{m_y} = \frac{z - z_B}{m_z} \] Since point B's coordinates are \((1,2,3)\) and the direction ratios are \((4,2,-4)\), the vector equation of line segment BC is: \[ \frac{x - 1}{4} = \frac{y - 2}{2} = \frac{z - 3}{-4} \]

Let P be the foot of the perpendicular from point A to the line BC. Let the parameter of the line be t. Then the coordinates of point P can be represented as: \( P(t) = (1 + 4t, 2 + 2t, 3 - 4t) \) Since AP is perpendicular to line segment BC, their dot product should be zero: \[ AP \cdot BC = 0 \] Dot product: \( (x_P - x_A)m_x + (y_P - y_A)m_y + (z_P - z_A)m_z = 0 \) Substitute P(t) and direction ratios: \[ ((1 + 4t)- 0)4 + ((2+ 2t) - (-1))2+((3 - 4t)- 4)(-4) = 0 \] Now, solve for t: \[ 4(1 + 4t) + 2(3 + 2t) - 4(7 - 4t) = 0 \] \[ 16t + 12t + 16t = 52 \] \[ 44t = 52 \] \[ t = \frac{13}{11} \]

Now we need to find the coordinates of the foot of the perpendicular, P(t) when t = \(\frac{13}{11}\): \( P(t) = (1 + 4t, 2 + 2t, 3 - 4t) \) \[ P\left(\frac{13}{11}\right) = \left(1 + 4\left(\frac{13}{11}\right), 2 + 2\left(\frac{13}{11}\right), 3 - 4\left(\frac{13}{11}\right)\right) \] \[ P\left(\frac{13}{11}\right) = \left(\frac{1+52}{11}, \frac{2+26}{11}, \frac{3-52}{11}\right) \] \[ P\left(\frac{13}{11}\right) = \left(\frac{53}{11}, \frac{28}{11}, \frac{-49}{11}\right) \] Thus, the coordinates of the foot of the perpendicular from point A on line segment BC are \(\frac{53}{11}\), \(\frac{28}{11}\), and \(\frac{-49}{11}\). Now, let's check which of the given options matches our calculated coordinates. Comparing with given options and rounding, none of the options match our calculated coordinates(\(\frac{53}{11}\), \(\frac{28}{11}\), \(\frac{-49}{11}\)). NOTE: It appears an error happened with the options provided in the exercise. So, the answer to this question should be stated as "none of the given options are correct".