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The direction vector of the first line, $\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$, is $\vec{b_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$. The direction vector of the second line, $\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu(3\hat{i}+8\hat{j}-5\hat{k})$, is $\vec{b_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.
Since the required line is perpendicular to both given lines, its direction vector $\vec{b}$ is parallel to the cross product of $\vec{b_1}$ and $\vec{b_2}$. $$\vec{b} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$$ $$\vec{b} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))$$ $$\vec{b} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)$$ $$\vec{b} = 24\hat{i} + 36\hat{j} + 72\hat{k}$$ We can simplify this vector by dividing by the common factor 12: $$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$$
The line passes through the point $(1, 2, -4)$, so the position vector of a point on the line is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$. The vector equation of the line is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter. $$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k})$$
The cartesian equation of the line passing through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is given by: $$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$ In this case, $(x_1, y_1, z_1) = (1, 2, -4)$ and the direction ratios are $2, 3, 6$. Therefore, the cartesian equation of the line is: $$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$$
Final Answer: Vector equation: $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k})$ Cartesian equation: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$