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(B) The equation of the plane containing the two lines is given by the determinant form: $\left| \begin{array}{ccc} x+2 & y-2 & z+5 \ 3 & 5 & 7 \ 1

Vedclass · Accepted Answer

$11/\sqrt{6}$

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(B) The equation of the plane containing the two lines is given by the determinant form: $\left| \begin{array}{ccc} x+2 & y-2 & z+5 \ 3 & 5 & 7 \ 1 & 4 & 7 \end{array} ight| = 0$ Expanding the determinant: $(x+2)(35-28) - (y-2)(21-7) + (z+5)(12-5) = 0$ $7(x+2) - 14(y-2) + 7(z+5) = 0$ Dividing by $7$: $(x+2) - 2(y-2) + (z+5) = 0$ $x - 2y + z + 2 + 4 + 5 = 0$ $x - 2y + z + 11 = 0$ The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. Here,$A=1, B=-2, C=1, D=11$. $d = \frac{|11|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{11}{\sqrt{1+4+1}} = \frac{11}{\sqrt{6}}$.

The perpendicular distance from the origin to the plane containing the two lines,$\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$,is

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