Direction cosine of line \(2 \mathrm{x}=3 \mathrm{y}+5, \mathrm{z}=7-(\mathrm{y} / 5)\) is (A) \([10 / \sqrt{(235)}],[15 / \sqrt{(235)}]\), [3 / V(235)] (B) \([(-10) / \sqrt{(} 235)],[(-15) / \sqrt{(235)}],[(-3) / \sqrt{(235)}]\) (C) \([10 / \sqrt{(235)}],[(-15) / \sqrt{(235)}],[3 / \sqrt{(235)}]\) (D) None of these

The given equations are \(2x = 3y + 5\) and \(z = 7 - \frac{y}{5}\). We can write the equations in terms of x, y, and z as follows: \(2x - 3y = 5\) and \(z + \frac{y}{5} = 7\) Now let's express y in terms of x and z by solving the first equation for y: \(y = \frac{2x - 5}{3}\) Now substitute this expression for y in the second equation: \(z + \frac{(2x - 5)}{15} = 7\)

Now that we have a single equation involving x, y, and z, we can find the direction ratios of the line. To do this, we'll take the coefficients of x, y, and z in the above equation: \(2x - 3(\frac{2x - 5}{3}) = 5\) Now multiply through by 3 to make the coefficients integers: \(6x - (2x - 5) = 15\) So the direction ratios of the line are 6, -2, and -1.

Now we'll normalize the direction ratios to find the direction cosines. To do this, we first find the magnitude of the direction ratios, and then divide each direction ratio by this magnitude: \(magnitude = \sqrt{6^2 + (-2)^2 + (-1)^2} = \sqrt{36 + 4 + 1} = \sqrt{41}\) Now divide each direction ratio by the magnitude: \(direction\ cosine\ of\ x = \frac{6}{\sqrt{41}}\) \(direction\ cosine\ of\ y = \frac{-2}{\sqrt{41}}\) \(direction\ cosine\ of\ z = \frac{-1}{\sqrt{41}}\) Now compare our calculated direction cosines with the given options: (A) \([\frac{10}{\sqrt{235}}, \frac{15}{\sqrt{235}}, \frac{3}{\sqrt{235}}]\) (B) \([\frac{-10}{\sqrt{235}}, \frac{-15}{\sqrt{235}}, \frac{-3}{\sqrt{235}}]\) (C) \([\frac{10}{\sqrt{235}}, \frac{-15}{\sqrt{235}}, \frac{3}{\sqrt{235}}]\) (D) None of these Our calculated direction cosines do not match with any of the given options. Thus, the correct answer is: (D) None of these