A. \( \overrightarrow n = (3; - 1;2).\)
B. \(\overrightarrow n = (3;0;2).\)
C. \(\overrightarrow n = ( - 1; - 1;2).\)
D. \(\overrightarrow n = (0; - 3;1).\)
A. \(D\left( { - 2;4; - 5} \right)\)
B. \(D\left( {6;2; - 3} \right)\)
C. \(D\left( {4;2;9} \right)\)
D. \(D\left( { - 4; - 2;9} \right)\)
A. \(I( - 4;4;6).\)
B. \(I(1;2;3).\)
C. \(I( - 1;2;3).\)
D. \(I(2;2;3).\)
A. \(\Delta :\frac{{x - 1}}{1} = \frac{{y + 3}}{1} = \frac{{z - 4}}{{ - 2}}\)
B. \(\Delta :\frac{{x - 1}}{1} = \frac{{y + 3}}{{ - 1}} = \frac{{z - 4}}{2}\)
C. \(\Delta :\frac{{x - 1}}{1} = \frac{{y + 3}}{{ - 1}} = \frac{{z - 4}}{{ - 2}}\)
D. \(\Delta :\frac{{x - 1}}{{ - 1}} = \frac{{y + 3}}{{ - 1}} = \frac{{z - 4}}{{ - 2}}\)
A. 1
B. 3
C. \(\sqrt 2 \)
D. 2
A. \({x^2} + {y^2} + {z^2} - 10x - 8y - 6z - 32 = 0\)
B. \({x^2} + {y^2} + {z^2} - 10x - 8y - 6z - 12 = 0\)
C. \({x^2} + {y^2} + {z^2} - 10x - 8y - 6z + 16 = 0\)
D. \({x^2} + {y^2} + {z^2} - 10x - 8y - 6z + 38 = 0\)
A. \(2x+3y=0\)
B. \(y+2z=0\)
C. \(2x-3y=0\)
D. \(3x+2y=0\)
A. \(\sqrt {65} \)
B. \(5\sqrt 3 \)
C. \(\sqrt {17} \)
D. \(5\sqrt 2 \)
A.
\(\left\{ \begin{array}{l}
x = - 1 + 2t\\
y = 2 - 2t\\
z = - 3{\rm{ }} - 3t
\end{array} \right.\)
B.
\(\left\{ \begin{array}{l}
x = - 1 + 2t\\
y = 2 + 2t\\
z = 3{\rm{ }} + 3t
\end{array} \right.\)
C.
\(\left\{ \begin{array}{l}
x = - 1 + 2t\\
y = 2 + 2t\\
z = - 3{\rm{ }} - 3t
\end{array} \right.\)
D.
\(\left\{ \begin{array}{l}
x = - 1 + 2t\\
y = 2 + 2t\\
z = - 3{\rm{ }} + 3t
\end{array} \right.\)
A.
\(d:\left\{ \begin{array}{l}
x = 1 + 2t\\
y = 2 - t\\
z = - 3 - 7t
\end{array} \right.\)
B.
\(d:\left\{ \begin{array}{l}
x = 1 + 2t\\
y = 2 + t\\
z = - 3 - 7t
\end{array} \right.\)
C.
\(d:\left\{ \begin{array}{l}
x = 1 + 3t\\
y = 2 - t\\
z = - 3 + t
\end{array} \right.\)
D.
\(d:\left\{ \begin{array}{l}
x = 1 + 4t\\
y = 2 + t\\
z = - 3 + t
\end{array} \right.\)
A.
\(\left\{ \begin{array}{l}
x = - 1 + 2t\\
y = 2\\
z = 3 + t
\end{array} \right.\)
B.
\(\left\{ \begin{array}{l}
x = 1 - t\\
y = 2\\
z = 3 + t
\end{array} \right.\)
C.
\(\left\{ \begin{array}{l}
x = - 1 - 2t\\
y = 2\\
z = 3 + t
\end{array} \right.\)
D.
\(\left\{ \begin{array}{l}
x = - 1 + t\\
y = 2\\
z = 3 - t
\end{array} \right.\)
A.
\(d:\left\{ \begin{array}{l}
x = 1 + t\\
y = 2 + t\\
z = 3 - 2t
\end{array} \right.\)
B.
\(d:\left\{ \begin{array}{l}
x = 1 - t\\
y = 2 + t\\
z = 3 - 2t
\end{array} \right.\)
C.
\(d:\left\{ \begin{array}{l}
x = 1 + t\\
y = 2 - t\\
z = 3 - t
\end{array} \right.\)
D.
\(d:\left\{ \begin{array}{l}
x = 2 - t\\
y = 1 + t\\
z = 1 + 2t
\end{array} \right.\)
A. \( - 5y + z + 3 - 4\sqrt {26} = 0, - 5y + z + 3 + 4\sqrt {26} = 0\)
B. \( - 5y + z + 3 - 4\sqrt {23} = 0, - 5y + z + 3 + 4\sqrt {23} = 0\)
C. \(5x + y + 3 - 4\sqrt {26} = 0,5x + y + 3 + 4\sqrt {26} = 0\)
D. \(5x - z + 3 - 4\sqrt {23} = 0,5x - z + 3 + 4\sqrt {23} = 0\)
A. \(\frac{{x + 2}}{1} = \frac{y}{{ - 1}} = \frac{{z - 3}}{1}\)
B. \(\frac{{x - 2}}{{ - 1}} = \frac{{y - 1}}{1} = \frac{z}{1}\)
C. \(\frac{{x - 2}}{{ - 1}} = \frac{y}{1} = \frac{{z + 3}}{{ - 1}}\)
D. \(x-2=y=z+3\)
A. m = 3
B. m = 2
C. m = 4
D. m = 1
A. \(2x + 2y - z - 6 = 0\)
B. \(2x + 2y - z - 2 = 0\)
C. \(2x + 2y - z + 2 = 0\)
D. \(2x + 2y - z - 6 = 0\)
A. \(\left( \alpha \right):x - 4z + 2 = 0\)
B. \(\left( \alpha \right):x + 4z - 1 = 0\)
C. \(\left( \alpha \right):4x - z + 1 = 0\)
D. \(\left( \alpha \right):2x + z - 3 = 0\)
A. \(V = \frac{{27}}{8}\)
B. \(V = \frac{{81\sqrt 3 }}{8}\)
C. \(V = \frac{{9\sqrt 3 }}{2}\)
D. \(V = \frac{{64}}{{27}}\)
A. \({x^2} + {y^2} + {\left( {z + 2} \right)^2} = 25\)
B. \({x^2} + {y^2} + {\left( {z + 2} \right)^2} = 16\)
C. \({x^2} + {y^2} + {\left( {z - 2} \right)^2} = 25\)
D. \({x^2} + {y^2} + {\left( {z - 2} \right)^2} = 16\)
A. \(2x + y - z + 8 = 0\)
B. \(5x - 11y - 3z + 1 = 0\)
C. \(2x - y - z + 2 = 0\)
D. \(3x - 2y - 4z + 1 = 0\)
A. \(x+z=1\)
B. \(y+z+3=0\)
C. \(x+z+1=0\)
D. \(x+y+1=0\)
A. \({x^2} + {\left( {y - 3} \right)^2} + {\left( {z + 1} \right)^2} = 3\)
B. \({x^2} + {\left( {y - 3} \right)^2} + {\left( {z + 1} \right)^2} = 9\)
C. \({x^2} + {\left( {y - 3} \right)^2} + {\left( {z - 1} \right)^2} = 9\)
D. \({x^2} + {\left( {y + 3} \right)^2} + {\left( {z - 1} \right)^2} = 9\)
A. \(b - 3c = 1\)
B. \(2b + c = 1\)
C. \(3b + c = 3\)
D. \(b + c = 1\)
A. \(\left( \alpha \right):\frac{x}{8} + \frac{y}{{ - 2}} + \frac{z}{4} = 0\)
B. \(\left( \alpha \right):\frac{x}{4} + \frac{y}{{ - 1}} + \frac{z}{2} = 1\)
C. \(\left( \alpha \right):x - 4y + 2z - 8 = 0\)
D. \(\left( \alpha \right):x - 4y + 2z = 0\)
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