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(B) The equation of the plane containing the two lines is given by the determinant equation:$\left|\begin{array}{ccc} x+1 & y-1 & z-3 \ 6 & 7 & 8 \

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(B) The equation of the plane containing the two lines is given by the determinant equation:$\left|\begin{array}{ccc} x+1 & y-1 & z-3 \ 6 & 7 & 8 \ 3 & 5 & 7 \end{array}ight| = 0$Expanding the determinant:$(x+1)(49-40) - (y-1)(42-24) + (z-3)(30-21) = 0$$9(x+1) - 18(y-1) + 9(z-3) = 0$Dividing by $9$,we get $x+1 - 2(y-1) + z-3 = 0$,which simplifies to $x - 2y + z = 0$.The length of the perpendicular $PQ$ from point $P(7, -2, 13)$ to the plane $x - 2y + z = 0$ is given by the formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.$PQ = \frac{|1(7) - 2(-2) + 1(13)|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|7 + 4 + 13|}{\sqrt{1 + 4 + 1}} = \frac{24}{\sqrt{6}} = 4\sqrt{6}$.Therefore,$(PQ)^2 = (4\sqrt{6})^2 = 16 	imes 6 = 96$.

Let $Q$ be the foot of the perpendicular from the point $P(7, -2, 13)$ on the plane containing the lines $\frac{x+1}{6} = \frac{y-1}{7} = \frac{z-3}{8}$ and $\frac{x-1}{3} = \frac{y-2}{5} = \frac{z-3}{7}$. Then $(PQ)^{2}$ is equal to ..... .

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