If lines \([(\mathrm{x}-1) / 2]=[(\mathrm{y}-3) / 4]=\mathrm{z}\) and \([(x-4) / 3]=[(1-y) / 2]=[(z-1) / 1]\) are co-planer, then the equation of plane containing these two lines is (A) \(6 \mathrm{x}+\mathrm{y}+16 \mathrm{z}=9\) (B) \(6 \mathrm{x}+\mathrm{y}-16 \mathrm{z}=9\) (C) \(6 \mathrm{x}-\mathrm{y}-16 \mathrm{z}=9\) (D) \(6 \mathrm{x}-\mathrm{y}+16 \mathrm{z}=9\)

For the first line, we have the symmetric equation: \[(\mathrm{x}-1) / 2 = (\mathrm{y}-3) / 4 = \mathrm{z}\] Let us assign a parameter \(t\) for the first line: \[t = (\mathrm{x}-1) / 2 = (\mathrm{y}-3) / 4 = \mathrm{z}\] Solving for x, y, and z, we get the parametric equation of the first line: \[\mathrm{x} = 2t + 1, \mathrm{y} = 4t + 3, \mathrm{z} = t\] Hence, a point on the first line, \(P_1\), can be obtained by taking \(t = 0\), and the direction vector , \(d_1\), by taking the derivative with respect to \(t\): \[P_1 = (1, 3, 0), d_1 = (2, 4, 1)\] Similarly, for the second line, we have the symmetric equation: \[((\mathrm{x}-4) / 3)=((1-\mathrm{y}) / 2)=((\mathrm{z}-1) / 1)\] Let us assign a parameter \(s\) for the second line: \[s = (\mathrm{x}-4) / 3 = (1-\mathrm{y}) / 2 = (\mathrm{z}-1) / 1\] Solving for x, y, and z, we get the parametric equation of the second line: \[\mathrm{x} = 3s + 4, \mathrm{y} = 1 - 2s, \mathrm{z} = s + 1\] Hence, a point on the second line, \(P_2\), can be obtained by taking \(s = 0\), and the direction vector , \(d_2\), by taking the derivative with respect to \(s\): \[P_2 = (4, 1, 1), d_2 = (3, -2, 1)\]

Since the lines are coplanar, their direction vectors, \(d_1\) and \(d_2\), should be parallel to the plane. We can find the normal vector of the plane, \(\mathrm{n}\), by taking the cross product of \(d_1\) and \(d_2\): \[\vec{n} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 1 \\ 3 & -2 & 1 \end{vmatrix} = (6, 1, -16) \]

Now, to find the equation of the plane, we can use the point-normal equation, where \(\vec{n} = (A, B, C)\) is the normal vector and \(P_1 = (x_1, y_1, z_1)\) is the point on the plane: \[A(x - x_1) + B(y - y_1) + C(z - z_1) = 0\] Substituting the values for \(\vec{n}\) and \(P_1\), we get: \[6(x - 1) + (y - 3) -16(z - 0) = 0 \Rightarrow 6x + y - 16z = 9\] Hence, the equation of the plane containing the given lines is: \(6x + y - 16z = 9\) which corresponds to option (A).