If the line $\frac{x}{2} = \frac{y}{- 3} = z$ is vertical, then the direction ratios of the line of greatest slope in the plane
$3 x - 2 y + z = 0$ are

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$(1, 1, 1)$

$(3, - 1, 1)$

$(11, 16, - 1)$

$(11, 16, 1)$

answer is C.

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Detailed Solution

Let $\vec{a}$ be vertical $\vec{b}$ be normal to the given plane.

The line of greatest slope on the given plane is
perpendicular to $\vec{b}$ and the line of intersection of the
two planes.

$∴ \vec{c} is along (\vec{a} \times \vec{b}) \times \vec{b}, \vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + \hat{k}$

$\vec{a} \times \vec{b} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & 1 \\ 3 & - 2 & 1 \end{array}| = - \hat{i} + \hat{j} + 5 \hat{k}$

$(\vec{a} \times \vec{b}) \times \vec{b} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & 5 \\ 3 & - 2 & 1 \end{array}| = 11 \hat{i} + 16 \hat{j} - \hat{k}$