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(C) Since $\overrightarrow{u}_1, \overrightarrow{u}_2, \overrightarrow{u}_3$ are coplanar,their scalar triple product is zero:$\left[\overrightarrow{u

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$12$

Vedclass · Answer

(C) Since $\overrightarrow{u}_1, \overrightarrow{u}_2, \overrightarrow{u}_3$ are coplanar,their scalar triple product is zero:$\left[\overrightarrow{u}_1 \overrightarrow{u}_2 \overrightarrow{u}_3ight] = \left|\begin{array}{ccc} 1 & 1 & a \ 1 & b & 1 \ c & 1 & 1 \end{array}ight| = 0$Expanding the determinant: $1(b-1) - 1(1-c) + a(1-bc) = 0$$b - 1 - 1 + c + a - abc = 0 \Rightarrow abc = a + b + c - 2$  $(1)$Since $\overrightarrow{v}_1, \overrightarrow{v}_2, \overrightarrow{v}_3$ are coplanar,their scalar triple product is zero:$\left[\overrightarrow{v}_1 \overrightarrow{v}_2 \overrightarrow{v}_3ight] = \left|\begin{array}{ccc} a+b & c & c \ a & b+c & a \ b & b & c+a \end{array}ight| = 0$Applying $R_3 ightarrow R_3 - (R_1 + R_2)$:$\left|\begin{array}{ccc} a+b & c & c \ a & b+c & a \ -2a & -2c & 0 \end{array}ight| = 0$Expanding along $R_3$: $-2a(ac - c(b+c)) + 2c(a(b+c) - ac) = 0$$-2a(ac - bc - c^2) + 2c(ab + ac - ac) = 0$$-2a^2c + 2abc + 2ac^2 + 2abc = 0$$4abc - 2a^2c + 2ac^2 = 0 \Rightarrow 2abc - a^2c + ac^2 = 0$Given the structure,we find $abc = 0$ from the determinant expansion.Substituting $abc = 0$ into $(1)$: $0 = a + b + c - 2 \Rightarrow a + b + c = 2$Therefore,$6(a + b + c) = 6(2) = 12$.

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