The square of the distance of the point (10/7, 22/7, 7) from the line (x+1)/3 = (y+3)/5 = (z+5)/7 in the direction of the vector i + 4j + 7k is:

Three Dimensional Geometry: JEE Main 2025 Mathematics question

JEE Main 2025 · January, Unspecified (session-level dataset) · Q92

JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumThree Dimensional Geometry249× since 2002

The square of the distance of the point

\left(\frac{10}{7}, \frac{22}{7}, 7\right)

from the line

\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}

in the direction of the vector

\hat{i} + 4\hat{j} + 7\hat{k}

is:

JEE Main 2025 Three Dimensional Geometry Question with Solution (January Unspecified (session-level dataset), Q92) | JEE Arc

JEE Main 2025 · January, Unspecified (session-level dataset) · Q92

JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumThree Dimensional Geometry249× since 2002

The square of the distance of the point

\left(\frac{10}{7}, \frac{22}{7}, 7\right)

from the line

\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}

in the direction of the vector

\hat{i} + 4\hat{j} + 7\hat{k}

is: