Distance of the point (2, 4, -1) from the line $\frac{10+2x}{2}=\frac{y+3}{4}=\frac{6-z}{9}$ is

7 units

The correct answer is Option (3) → 7 units

Given point: \(P(2, 4, -1)\)

Given line:

\(\frac{10 + 2x}{2} = \frac{y + 3}{4} = \frac{6 - z}{9} = t\)

Rewrite parametric form:

\[ \begin{cases} \frac{10 + 2x}{2} = t \Rightarrow 10 + 2x = 2t \Rightarrow x = t - 5 \\ \frac{y + 3}{4} = t \Rightarrow y = 4t - 3 \\ \frac{6 - z}{9} = t \Rightarrow 6 - z = 9t \Rightarrow z = 6 - 9t \end{cases} \]

Direction vector of line:

\(\vec{d} = (1, 4, -9)\)

Point on line at \(t=0\):

\(Q = (-5, -3, 6)\)

Vector \( \vec{PQ} = \vec{P} - \vec{Q} = (2 + 5, 4 + 3, -1 - 6) = (7, 7, -7)\)

Distance from point to line:

\[ \text{Distance} = \frac{|\vec{PQ} \times \vec{d}|}{|\vec{d}|} \]

Calculate \(\vec{PQ} \times \vec{d}\):

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 7 & 7 & -7 \\ 1 & 4 & -9 \\ \end{vmatrix} = \mathbf{i}(7 \cdot (-9) - (-7) \cdot 4) - \mathbf{j}(7 \cdot (-9) - (-7) \cdot 1) + \mathbf{k}(7 \cdot 4 - 7 \cdot 1) \]

\[ = \mathbf{i}(-63 + 28) - \mathbf{j}(-63 + 7) + \mathbf{k}(28 - 7) = \mathbf{i}(-35) - \mathbf{j}(-56) + \mathbf{k}(21) = (-35, 56, 21) \]

Magnitude of cross product:

\[ | \vec{PQ} \times \vec{d} | = \sqrt{(-35)^2 + 56^2 + 21^2} = \sqrt{1225 + 3136 + 441} = \sqrt{4802} \]

Magnitude of \(\vec{d}\):

\[ |\vec{d}| = \sqrt{1^2 + 4^2 + (-9)^2} = \sqrt{1 + 16 + 81} = \sqrt{98} \]

Distance:

\[ = \frac{\sqrt{4802}}{\sqrt{98}} = \sqrt{\frac{4802}{98}} = \sqrt{49} = 7 \]