<p>The point of intersection of the plane \(\vec{r} \cdot (3\hat{i} - 5\hat{j} + 2\hat{k}) = 6\) with the straight line passing through the origin and perpendicular to the plane \(2x - y - z = 4\), is \((x_0, y_0, z_0)\). The value of \((2x_0 - 3y_0 + z_0)\) is:</p>
Step-by-Step Solution
Key Concept: The line perpendicular to plane 2x - y - z = 4 has direction vector equal to the plane's normal (2, -1, -1). Parametrize this line through origin as r(t) = t(2, -1, -1) and substitute into the given plane equation to find t, then extract coordinates.
Step 1: Find the direction vector of the line perpendicular to plane 2x - y - z = 4. The normal vector to this plane is n = (2, -1, -1), which is the direction of the perpendicular line. Step 2: Write the parametric equation of the line passing through origin with direction (2, -1, -1): r(t) = t(2, -1, -1) = (2t, -t, -t) Step 3: Substitute this into the plane equation $\vec{r} \cdot (3\hat{i} - 5\hat{j} + 2\hat{k}) = 6$: (2t, -t, -t) · (3, -5, 2) = 6 6t + 5t - 2t = 6 9t = 6 t = 2/3 Step 4: Find the point of intersection $(x_0, y_0, z_0)$: $x_0 = 2t = 2(2/3) = 4/3$ $y_0 = -t = -2/3$ $z_0 = -t = -2/3$ Step 5: Calculate $2x_0 - 3y_0 + z_0$: $= 2(4/3) - 3(-2/3) + (-2/3)$ $= 8/3 + 2 - 2/3$ $= 8/3 - 2/3 + 2$ $= 6/3 + 2 = 2 + 2 = 4$ ∴ Answer: 4
Correct Answer: D