If a, b, c are the intercepts made on X, Y, Z | vedclass

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(C) The equation of a plane passing through three points $(x_1, y_1, z_1), (x_2, y_2, z_2)$ and $(x_3, y_3, z_3)$ is given by the determinant equation

Vedclass · Accepted Answer

-$15$

Vedclass · Answer

(C) The equation of a plane passing through three points $(x_1, y_1, z_1), (x_2, y_2, z_2)$ and $(x_3, y_3, z_3)$ is given by the determinant equation: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$ Substituting the given points $(1, 0, -2), (3, -1, 2)$ and $(0, -3, 4)$: $\begin{vmatrix} x-1 & y-0 & z+2 \ 3-1 & -1-0 & 2+2 \ 0-1 & -3-0 & 4+2 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-1 & y & z+2 \ 2 & -1 & 4 \ -1 & -3 & 6 \end{vmatrix} = 0$ Expanding the determinant: $(x-1)(-6 + 12) - y(12 + 4) + (z+2)(-6 - 1) = 0$ $6(x-1) - 16y - 7(z+2) = 0$ $6x - 6 - 16y - 7z - 14 = 0 \Rightarrow 6x - 16y - 7z = 20$ Dividing by $20$ to get the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$: $\frac{6x}{20} - \frac{16y}{20} - \frac{7z}{20} = 1 \Rightarrow \frac{x}{10/3} + \frac{y}{-20/16} + \frac{z}{-20/7} = 1$ Thus,$a = \frac{10}{3}, b = -\frac{5}{4}, c = -\frac{20}{7}$ Calculating $3a + 4b + 7c$: $3(\frac{10}{3}) + 4(-\frac{5}{4}) + 7(-\frac{20}{7}) = 10 - 5 - 20 = -15$.

If $a, b, c$ are the intercepts made on $X, Y, Z$-axes respectively by the plane passing through the points $(1, 0, -2), (3, -1, 2)$ and $(0, -3, 4)$,then $3a + 4b + 7c =$

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