Let P be a plane containing the line frac{x-1 | vedclass

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(C) The equation of a plane containing the line passing through $(x_1, y_1, z_1)$ with direction ratios $(a_1, b_1, c_1)$ and parallel to a line with

Vedclass · Accepted Answer

$38$

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(C) The equation of a plane containing the line passing through $(x_1, y_1, z_1)$ with direction ratios $(a_1, b_1, c_1)$ and parallel to a line with direction ratios $(a_2, b_2, c_2)$ is given by the determinant equation:$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}ight| = 0$Substituting the given values $(x_1, y_1, z_1) = (1, -6, -5)$,$(a_1, b_1, c_1) = (3, 4, 2)$,and $(a_2, b_2, c_2) = (4, -3, 7)$:$\left|\begin{array}{ccc} x-1 & y+6 & z+5 \ 3 & 4 & 2 \ 4 & -3 & 7 \end{array}ight| = 0$Since the point $(1, -1, \alpha)$ lies on the plane $P$,we substitute $x=1, y=-1, z=\alpha$ into the determinant:$\left|\begin{array}{ccc} 1-1 & -1+6 & \alpha+5 \ 3 & 4 & 2 \ 4 & -3 & 7 \end{array}ight| = 0$$\left|\begin{array}{ccc} 0 & 5 & \alpha+5 \ 3 & 4 & 2 \ 4 & -3 & 7 \end{array}ight| = 0$Expanding along the first row:$0(28 - (-6)) - 5(21 - 8) + (\alpha+5)(-9 - 16) = 0$$-5(13) + (\alpha+5)(-25) = 0$$-65 - 25\alpha - 125 = 0$$-25\alpha - 190 = 0$$25\alpha = -190$$5\alpha = -38$Therefore,$|5\alpha| = |-38| = 38$.

Let $P$ be a plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane $P$,then the value of $|5\alpha|$ is equal to ....... .

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