The distance between the point (3, 4, 5) and the point where the line…

The correct answer is Option (1) → 3

Let

    \( \displaystyle \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \)

A general point on the line is \((\lambda + 3,\ 2\lambda + 4,\ 2\lambda + 5)\). It lies on the plane x + y + z = 17 when

    \((\lambda + 3) + (2\lambda + 4) + (2\lambda + 5) = 17\)
    \(5\lambda + 12 = 17\)
    \(\lambda = 1\)

So the line meets the plane at (4, 6, 7). Its distance from (3, 4, 5):

    \( \displaystyle d = \sqrt{(4-3)^2 + (6-4)^2 + (7-5)^2} \)

    \( \displaystyle = \sqrt{1 + 4 + 4} = 3 \)

Shortcut: (3, 4, 5) is the point with \(\lambda = 0\), and the direction ratios 1, 2, 2 have length \(\sqrt{1 + 4 + 4} = 3\). So the point with \(\lambda = 1\) is at distance \(1 \times 3 = 3\).

Note: This topic was not in the CUET 2026 syllabus.